LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


SHOP  PROBLEMS  IN 
MATHEMATICS 


BY 

WILLIAM  E.  BRECKENRIDGE,  M.A. 

CHAIRMAN  OF  THE  DEPARTMENTS  OF  MATHEMATICS  IN  THE  STUYVESANT 

HIGH  SCHOOL  AND  THE  STUYVESANT  EVENING  TRADE  SCHOOL,  NEW 

YORK  CITY  ;   DEPUTY  EXAMINER  IN  SHOP  MATHEMATICS  FOR 

THE   INTERNATIONAL  COMMITTEE   OF   THE   Y.  M.  C.  A. 


SAMUEL  F.  MERSEREAU,  B.A. 

CHAIRMAN  OF  THE  DEPARTMENT  OF  WOODWORKING, 

STUYVESANT  HIGH   SCHOOL^  NEW  YORK  CITY, 

PRACTICAL  CARPENTER  AND  BUILDER 


CHARLES  F.  MOORE,  B.S. 

CHAIRMAN  OF  THE  DEPARTMENT  OF   METAL  WORK,  STUYVESANT 
HIGH   SCHOOL  ;   INSTRUCTOR  IN    MACHINE-SHOP   PRAC- 
TICE IN  THE   STUYVESANT   EVENING  TRADE 
SCHOOL,   NEW  YORK    CITY 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1910,  BY 
WILLIAM    E.  BRECKENRIDGE,  SAMUEL  F.  MERSEREAU, 

AND  CHARLES  F.  MOORE 

ENTERED  AT  STATIONERS'  HALL 

ALL  RIGHTS  RESERVED 

710.6 


gtftengum 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PKEFACE 


The  aim  of  this  book  is  twofold :  first,  to  impart  to  the 
student  information  in  regard  to  shops  and  shop  materials, 
including  the  names  of  the  parts  of  machines  used  in  wood 
working  and  metal  working.;  second,  to  give  a  thorough 
training  in  the  mathematical  operations  that  are  useful  in 
shop  practice  and  science. 

A  book  of  this  size  cannot  be  a  treatise,  but  can  deal  only 
with  the  most  important  points  in  shop  work.  The  cuts  of 
the  lathes,  planers,  band  saw,  and  milling  machine  will  be 
found  useful  in  teaching  the  names  of  the  parts  of  these 
machines  and  in  illustrating  the  mathematical  problems.  To 
the  problems  sufficient  theory  has  been  added  to  make  the 
book  useful  as  a  textbook  in  courses  ^in  shop  mathematics. 
For  training  in  mathematical  operations  a  certain  amount 
of  drill  is  necessary  on  abstract  work  —  problems  without 
content  -+—  to  concentrate  attention  on  the  operations  and 
develop  speed  and  accuracy  in  computation.  For  this  pur- 
pose the  latter  part  of  the  book  is  devoted  to  a  Review  of 
Calculation.  It  is  desirable  that  some  of  this  abstract  work 
should  be  studied  every  day  in  connection  with  the  concrete 
problems.  The  slide  rule  has  been  treated  at  length  because 
the  authors  believe  that  it  should  be  in  more  common  use. 
Emphasis  has  been  laid  on  short  methods  and  on  checks. 
In  the  chapter  on  Ratio  and  Proportion  athletic  fields  have 
been  used  as  interesting  subjects  of  work. 

The  scope  of  the  work  on  the  informational  side  may  be 
seen  from  the  Table  of  Contents.  On  the  mathematical  side 


208529 


iv  SHOP  PROBLEMS  IN  MATHEMATICS 

the  problems  vary  from  the  addition  of  fractions,  which 
must  be  reviewed  by  all  men  who  enter  trade  schools,  to 
the  trigonometry  that  is  used  in  the  more  difficult  parts  of 
machine  work. 

The  arrangement  is  by  subjects  in  order  that  a  student 
may  select  those  problems  in  which  he  is  most  interested, 
and  that  a  teacher  may  readily  arrange  a  course  to  meet  the 
needs  of  his  particular  school.  It  is  suggested  that  the  work 
of  the  shop  be  correlated  with  that  of  the  mathematical 
classroom.  For  example,  when  a  student  in  a  manual  train- 
ing high  school  is  beginning  to  handle  boards  in  the  shop, 
he  is  commencing  the  study  of  algebra  in  his  mathematical 
classroom.  Let  him  be  assigned  some  problems  in  the  chap- 
ter on  Board  Measure,  together  with  a  review  of  work  in 
fractions.  From  these  problems  he  will  acquire  consider- 
able information  that  will  be  useful  in  the  shop,  and  from 
the  formulas  in  board  measure  he  will  understand  the 
meaning  of  letters  as  symbols  of  general  number.  If  I  and 
t  stand  for  the  length  and  thickness  of  the  board  that  the 
student  has  just  sawed  off  in  the  shop,  they  will  have  a 
definite  meaning  to  him. 

The  authors  hope  that  the  book  will  be  found  useful  in 
any  school  where  there  are  shops.  It  is  adapted  for  use  in 
the  shop  or  the  mathematical  classroom,  or  in  both. 

A  student  of  this  book  should  develop  an  appreciation  of 
the  value  of  mathematics  in  practical  work,  since  all  prob- 
lems in  the  book  are  based  on  actual  shop  practice.  He 
should  learn  to  make  a  formula,  solve  it  for  any  letter,  and 
apply  it  intelligently.  As  he  progresses  the  approximate 
nature  of  the  results  in  real  problems  should  become  evi- 
dent. He  should  learn  that  real  problems  very  rarely  "come 
out  even," —  that  there  is  usually  an  allowance  necessary  for 
waste,  and  he  should  develop  a  certain  judgment  as  to  what 
that  allowance  is.  He  should  find  that  different  degrees  of 


PREFACE  v 

accuracy  are  desirable  in  different  kinds  of  work,  —  a  result 
correct  to  "feet  being  sufficiently  accurate  in  certain  pieces 
of  work,  while  in  others,  measurements  are  required  to  ten- 
thousandths  of  an  inch. 

In  some  cases  where  it  seems  desirable  for  the  student  to 
become  familiar  with  several  systems  of  notation  in  common 
use,  different  forms  have  been  used  in  different  parts  of  the 
book ;  for  example,  3  ft.  6  in.  and  3'  6". 

The  book  is  designed  to  correlate  the  work  of  the  mathe- 
matical classroom  with  that  of  the  departments  of  mechanic 
arts  and  science.  This  correlation  is  impossible  without  the 
help  of  the  teachers  in  these  departments.  Their  cordial 
cooperation  in  this  school  is  hereby  acknowledged.  The 
authors  especially  acknowledge  their  indebtedness  to  Dr. 
Ernest  R.  von  Nardroff  for  valuable  suggestions,  and  to  Mr. 
F.  H.  Law  for  corrections  in  English. 


CONTENTS 


CHAPTER  PAGE 

I.    BOARD  MEASURE 1 

II.    HOUSE  BUILDING,  GENERAL  CONSTRUCTION,  HEIGHTS 

OF  TREES  AND  OTHER  MEASUREMENTS 17 

III.  PULLEYS,  BELTS,  AND  SPEEDS 62 

IV.  AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS  ;  TURNED 

WORK 74 

V.    PATTERN  MAKING  AND  FOUNDRY  WORK  ;  WEIGHTS  OF 

CASTINGS  FROM  WEIGHTS  OF  PATTERNS 87 

VI.    LENGTH    OF    STOCK     FOR    FORCINGS,     STRENGTH    OF 

FORCINGS,  ALLOWANCE  FOR  SHRINK  FITS  ....  102 

VII.    SPEEDS  OF  PULLEYS,  SHAFTS,  AND  GEARS Ill 

VIII.    CUTTING  SPEED  AND  FEED 110 

IX.    MICROMETER,  VERNIER,  AND  TAPERS  . 133 

X.    THREAD    PROPORTIONS,    GEARING    FOR    SCREW    CUT- 
TING, INDEXING       147 

XI.    GEAR  PROPORTIONS  AND  SPIRALS 166 

XII.    THE  UNIVERSAL  GRINDER  AND  THE  GAS  ENGINE    .    .  183 

REVIEW  OF  CALCULATION  WITH   SHORT 
METHODS 

XIII.  FRACTIONS 186 

XIV.  PERCENTAGE,.    AVERAGES.    SQUARE  ROOT 193 

XV.    RATIO  AND  PROPORTION 199 

XVI.    MENSURATION 213 

XVII.    FORMULAS .    . 237 

XVIII.    SOLUTION   OF  RIGHT  TRIANGLES   BY  NATURAL  FUNC- 
TIONS    249 

XIX.    SHORT  METHODS 253 

XX.    THE  SLIDE  RULE 260 

TABLES 274 

INDEX 279 

vii 


"€^R^JEI^ 

^     Of  THE 

UNIVERSITY 

OF 


SHOP  PROBLEMS 
MATHEMATICS 


CHAPTER  I 
BOARD  MEASURE 

1.  A  board  foot.  A  foot  in  board  measure,  or  a  board  foot, 
means  a  piece  of  lumber  having  an  area  of  1  sq.  ft.  on  its 
flat  surface  and  a  thickness  of  1  in.  or  less. 


U   12"  -4-    12."  -^     /2"J<-    /2"   *U  12"     J* 
FIG.  1.    BOARD  FEET 


2.  Feet.    The  term  "  feet  "  is  generally  used  for  "  board 
feet,"  except  in  places  where  it  is  likely  to  be  misunder- 
stood, when  "  board  feet,"  "  square  feet,"  "  linear  feet," 
etc.,  are  designated. 

3.  Rough    stock.     The   term    "  rough    stock  "   ordinarily 
means  lumber  roughly  gotten  out  for  some  purpose,  but 
having  its  dimensions  a  little  larger  than  is  actually  re- 
quired, to  allow  for  planing,  truing  up,  etc. 

4.  Dressed.    The  term  "  dressed  "  has  very  much  the  same 
meaning  as  "  planed,"  and  is  sometimes  used  in  connection 
with  a  piece  that  has  any  of  its  surfaces  planed,  as  "  dressed 
on  one   or   both  sides."    But  for   convenience  in  stating 

1 


2  SHOP  PROBLEMS  IN  MATHEMATICS 

problems,  etc.,  we  shall  use  the  term  to  designate  lumber 
planed  on  all  four  of  its  surfaces,  unless  otherwise  stated. 

5.  Allowance  for  dressing,  i.e.  planing.  If  lumber  is 
dressed,  it  loses  in  size  the  amount  taken  off  in  shav- 
ings. Usually  for  stock  1^  in.  or  more  in  thickness  the 
loss  is  about  |  in.  on  each  surface  planed.  Hence  a  piece 


PLANER 
(By  permission  of  the  American  Woodworking  Machinery  Co.) 

8  in.  wide  and  2  in.  thick  when  rough  becomes  7|  in.  wide 
and  If  in.  thick  when  dressed,  if  planed  on  all  four  surfaces. 

For  lumber  less  than  1^  in.  thick  the  amount  taken  off  in 
planing  is  usually  about  £  in.  for  both  surfaces.  This  varies, 
however,  according  to  the  size  and  nature  of  the  stock. 

In  ordinary  lumber  the  loss  of  size  due  to  planing  is  not 
taken  into  consideration  when  the  lumber  is  sold.  The  pur- 
chaser expects  to  lose  the  difference  between  the  rough  and 
dressed  lumber,  and  orders  enough  to  make  up  this  loss. 


BOARD  MEASURE  3 

6.  Width  of  rough  lumber.     In  measuring  the  width  of 
common  rough  lumber,  a  fraction  of  an  inch  equal  to  or 
greater  than  \  is  counted  as  a  whole  inch,  while  a  fraction 
less  than  1  is  neglected.     For  example,   a  board  6^  in. 
wide  would  be  called  7  in.  wide,  a  board  6f  in.  wide  would 
be  called  7  in.  wide,  and  one  G|  in.  wide  would  be  called 
6  in.  wide. 

7.  Surfaced.    The  term  "  surfaced  "  is  usually  applied  to 
boards  or  planks  that  are  planed  on  one  or  both  sides. 

8.  Jointed.    The  term  "jointed"  has  reference  to  lumber 
planed  on  its  edges.    It  is  also  used  to  designate  pieces  that 
are  made  straight  on  the  edges. 

9.  Allowance  in  cutting  logs.    Logs  are  supposed  to  be  cut 
a  few  inches  longer  than  the  length  actually  required,  to 
allow  for  bruising  and  other  damage  done  to  the  ends  in 
the  lumbering  operations.    The  amount  allowed  is  usually 
sufficient  to  permit  of  squaring  the  ends  of  the  lumber  and 
still  have  the  pieces  long  enough.    However,  it  is  well  to 
make  sure  of  the  lengths  when  purchasing,  or.  to  make  a 
small  allowance,  when  planning  certain  kinds  of  work,  by 
reducing  the  dimensions  a  trifle. 

10.  Lumber.    The  term  "  lumber  "  is  generally  applied  to 
pieces  not  more  than  4  in.  the  smaller  way,  usually  thickness. 

11.  Timber.    The  term  "timber"  is  applied  to  pieces  more 
than  4  in.  the  smaller  way.    It  is  also  applied  to  trees  stand- 
ing in  the  forest,  as  "one  hundred  acres  of  timber."   It  is 
sometimes  used  in  speaking  of  the  quality  of  wood,  as  "  a 
piece  contains  good  timber." 

12.  Board  and  plank.    Any  piece  of  lumber  under  1^  in. 
thick  is  usually  called  a  "  board  " ;  any  piece  from  1^  in. 
to  4  in.  thick  is  called  a  "  plank."    The  use  of  these  terms 
differs,  however,  in  various  localities. 

13.  Rule  for  board  feet.    From  section  1  it  is  evident  that 
the  rule  for  calculating  board  feet  is  as  follows : 


SHOP  PROBLEMS  IN  MATHEMATICS 


To  find  the  number  of  board  feet  in  a  piece  of  lumber, 
multiply  the  number  of  square  feet  in  its  fiat  surface  by  the 
number  of  inches  in  thickness,  counting  a  thickness  less  than 
one  inch  as  an  inch. 

EXAMPLE  1.  Find  the  number  of  board  feet  in  a  piece  of 
lumber  1  in.  thick,  9  in.  wide,  and  14  ft.  long. 

~^  x  1  -  10i  ft. 

EXAMPLE  2.  Find  the  number  of  board  feet  in  a  piece  of 
lumber  i  in.  thick,  8  in.  wide,  and  12  ft.  long. 


8x12 
12 


X  1  =  8  f  t. 


\  in.  in  thickness  is  counted 
as  1  in. 

EXAMPLE  3.  Find  the 
number  of  board  feet  in 
a  piece  of  timber  7  in. 
thick,  8  in.  wide,  and  18  ft. 
long.  (See  Fig.  2.) 


14.  Table  for  calculating  board  measure.  The  following 
simple  table  familiarized  will  be  found  very  useful  in  cal- 
culating board  measure  : 

FOR  LUMBER  1  IN.  OR  LESS  IN  THICKNESS 

Pieces  3  in.  wide  contain  £  as  many  feet  as  they  are  feet  long. 
Pieces  4  in.  wide  contain  1  as  many  feet  as  they  are  feet  long. 
Pieces  6  in.  wide  contain  |  as  many  feet  as  they  are  feet  long. 
Pieces  9  in.  wide  contain  f  as  many  feet  as  they  are  feet  long. 
Pieces  12  in.  wide  contain  as  many  feet  as  they  are  feet  long. 
Pieces  15  in.  wide  contain  \\  as  many  feet  as  they  are  feet  long. 
Pieces  16  in.  wide  contain  \\  as  many  feet  as  they  are  feet  long. 


BOARD  MEASURE  5 

ORAL    EXERCISE 

According  to  the  above  table  determine  orally  the  num- 
ber of  feet  in  pieces  having  the  following  dimensions  : 

1.  1  in.  by  3  in.  by  12  ft.  13.  8  in.  by  9  in.  by  20  ft. 

2.  1  in.  by  3  in.  by  16  ft.  14.  7  in.  by  9  in.  by  14  ft. 

3.  i- in.  by  4  in.  by  12  ft.  15.  12  in.  by  12  in.  by  12ft. 

4.  i  in.  by  4  in.  by  18  ft.  16.   14  in.  by  15  in.  by  16ft. 

5.  §  in.  by  4  in.  by  14  ft.  17.  7  in.  by  18  in.  by  20  ft. 

6.  2  in.  by  4  in.  by  12  ft.  18.  Two  pieces  each  3  in. 

7.  li  in.  by  3  in.  by  16  ft.  bJ  6  in.  by  16  ft. 

8.  li  in.  by  3  in.  by  16  ft.  19-  Two  pieces  each  4  in. 

9.  If  in.  by  4  in.  by  12  ft.  b?  6  in'  b^  14  ft' 

10.  5  in.  by  4  in.  by  18  ft.  20'  Two  Pieces  each  6  in' 

by  6  in.  by  13  ft. 

11.  1  in.  by  6m.  by  14  ft. 

12.  2  in.  by  9  in.  by  16  ft. 

15.  Reducing  inches  to  decimals  of  a  foot.    EXAMPLE.    Re- 
duce 2  ft.  5  in.  to  feet. 

2  ft.  5  in.  =  2^  ft.  =  2.4  ft. 
For  the  reduction  of  a  fraction  to  a  decimal,  see  sect.  162. 

EXERCISE 

Reduce  to  feet  and  decimals  of  a  foot  : 

1.  6  in.  5.   7  in.  9.  3  ft.  4  in. 

2.  3  in.  6.  11  in.  10.  4  ft.  10  in. 

3.  9  in.  7.  1  ft.  6  in.  11.  6  ft.  7  in. 

4.  5  in.  8.  2  ft.  3  in. 

16.  Boards  of  the  same  length  but  of  different  widths  (short 
method).    When  several  boards  of  the  same  length  but  of 


6  SHOP  PROBLEMS  IN  MATHEMATICS 

different  widths  are  to  be  measured,  it  is  sometimes  con- 
venient to  add  all  the  widths  together  and  figure  the  lot 
as  one  board.  A  good  way  to  add  these  is  to  run  off,  on  a 
tapeline,  the  width  of  each  board  separately  until  all  are 
measured  ;  the  last  figure  reached  on  the  line  will  be  the 
sum  of  the  widths  of  all  the  boards. 

17.  Standard  lengths  of  lumber.  In  selecting  lumber  it 
should  be  borne  in  mind  that  in  most  sections  the  standard 
lengths  are  10,  12,  14,  16,  18  ft.,  etc.  If  it  is  cut  to  a  spe- 
cial length,  it  always  costs  more. 

In  the  Adirondacks  one  of  the  standard  lengths  of  spruce 
is  13  ft. 

PROBLEMS 

1.  How  many  board  feet  in  a  piece  of  lumber  1  in.  thick, 
12  in.  wide,  and  5  ft.  long  ?    (See  Fig.  1.) 

2.  How  many  board  feet  in  a  piece  of  lumber  1^  in.  thick, 
11  in.  wide,  and  16  ft.  long  ?    (See  Fig.  3.) 


~\nj 

» 


v ' 

FIG.  3.    BOARD  MEASURE 

3.  A  piece  10  ft.  long,  2  in.  thick,  and  11  in.  wide  con- 
tains how  many  board  feet  ? 

4.  How  many  board  feet  of  stock  are  required  to  con- 
struct a  platform  8  ft.  6  in.  square,  if  the  stock  is  1^  in. 
thick  and  we  allow  three  board  feet  for  waste  in  squaring 
up  the  ends  of  the  boards  ? 

5.  A  piece  of  stock  is  ^  in.  thick,  9|  in.  wide,  and  16  ft. 
long.    How  many  board  feet  does  it  contain  ?    (See  sect.  6.) 

6.  How  many  board  feet  in  a  piece  18  ft.  long,  1\  in. 
wide,  and  \\  in.  thick  ?    (See  sect.  6.) 


BOARD  MEASURE  7 

7.  A  piece  of  lumber  is  10  in.  wide  at  one  end  and  8  in. 
wide  at  the  other.    If  it  is  12  ft.  long  and  1  in.  thick,  how 
many  board  feet  does  it  contain  ? 

8.  How  many  board  feet  in  a  stick  6  in.  thick,  8  in.  wide, 
and  14  ft.  long  ? 

9.  A  board  |  in.  thick,  8  in.  at  one  end,  12  in.  at  the 
other,  and  13^  ft.  long  contains  how  many  board  feet  ? 

10.  A  piece  of  2^-in.  plank  is  18  ft.  long,  9  in.  wide  at 
one  end,  and  12  in.  wide  at  the  other.    How  many  feet  does 
it  contain  ? 

11.  Five  pieces  of  lumber  (see  Fig.  4)  are  3,  4,  7,  9,  and 
5  in.  wide  respectively.  What  is  the  total  number  of  feet  in 
the  lot,  if  the  boards  are  1  in.  thick  and  10  ft.  long  ? 


t*M     ?  9 

FIG.  4.    SHOKT  METHOD  FOR  BOARDS  OF  DIFFERENT  WIDTHS 

12.  A  pile  contains  nine  boards  whose  widths  are  3,  4,  5, 
7,  9,  11,  13i,  15,  16^  in.  respectively.    If  the  boards  are  16 
ft.  long  and  we  saw  off  1  in.  on  each  end  of  every  board  for 
squaring,  how  many  square  feet  of  floor  will  the  boards 
cover  ? 

13.  A  stack  of  lumber  7  ft.  5  in.  wide  and  9  ft.  long  is 
composed  of  90  layers  of  boards  1  in.  thick,  placed  edge  to 
edge.    How  many  feet  does  the  stack  contain  ? 


8  SHOP  PROBLEMS  IN  MATHEMATICS 

14.  How  many  feet  in  a  wagon  load  of  lumber  14  ft. 
long,  3  ft.  2  in.  wide,  and  24  in.  high,  if  the  boards  are 
£  in.  thick  ? 

o 

15.  The  distance  between  the  bolster  stakes  of  a  lumber 
wagon  is  3  ft.  2  in.    If  we  put  on  a  load  of  planks,  dressed 
on  two  sides  only,  that  were  2  in.  thick  when  rough,  and 
the  load  is  16  ft.  long  and  35  in.  high,  how  many  board  feet 
does  the  load  contain  ? 

16.  The  back  stop  for  a  baseball  diamond  is  constructed 
of  9-in.  boards  I  in.  thick  and  16  ft.  long.    If  it  is  12  ft. 
high  and  16  ft.  long,  how  many  boards  does  it  require  ? 
How  many  board  feet  are  required  for  its  construction,  if 
we  have  3  uprights  18  ft.  long  and  3  braces  12  ft.  long,  the 
uprights  and  braces  to  be  2  in.  by  4  in.  ? 

17.  At  $35  per  M,  how  much  will  it  cost  for  lumber  for 
a  platform  20  ft.  wide  at  one  end,  25  ft.  wide  at  the  other, 
and  40  ft.  long,  the  lumber  to  be  2  in.  thick,  and  1  per  cent 
to  be  added  for  waste  in  squaring  off  the  ends  of  the  planks  ? 

18.  A  platform  12  ft.  wide  and  36  ft.  long  is  constructed 
of  If -in.  by  9|-in.  hemlock  planks  (dressed).    If  we  add  l£ 
per  cent  for  waste  in  squaring,  how  much  lumber  must  we 
buy  for  the  platform  ?    (See  sect.  5.) 

19.  The  foundation  for  the  floor  of  a  factory  building  is 
built  up  of  If-in.  by  5f-in.  by  12-ft.  dressed  hemlock  lum- 
ber spiked  together  sidewise  into  one  solid  piece  (see  Fig.  5). 
If  the  building  is  108  ft.  wide  and  145  ft.  10  in.  long, 
how  many  pieces  of  lumber  does  the  foundation  require, 
and  how   many  board  feet  do  they  contain,   making  no 
allowance  for  waste  ? 

20.  How   many   planks   (dressed)  are   required  to   con- 
struct a  solid  girder  49  ft.  long,  9|  in.  deep,  and  8|  in. 
wide,  if  the  planks  are  1|  in.  thick,  9|  in.  wide,  and  14 
ft.  long,  and  are   spiked  together  flatwise  ?    How  many 


BOARD  MEASURE 


9 


board  feet  does  the  girder  contain,  if  we  add  one  board  foot 
to  every  hundred  for  squaring  ?    (See  Fig.  5.) 

21.  At  $40  per  M,  how  much  will  enough  hard  maple 
lumber  cost  to  floor  a  bridge  16  ft.  wide  that  spans  a  river 
1000  ft.  wide,  the  planks  to  be  3  in.  thick  and  laid  at  an 
angle  of  45  degrees  (diagonally)  ?  Add  3  board  feet  to  every 
100  for  waste  in  cutting  the  planks  on  a  slant. 


FIG.  5.    FACTORY  FOUNDATION 

22.  The  sidewalk  on  the  bridge  in  Problem  21  is  6  ft. 
wide  and  is  laid  of  hard-pine  planks  If  in.  by  7|  in.  by 
12  ft.  long  (dressed).     How  many  planks  does  the  walk 
require,  and  how  many  feet  do  they  contain,  making  no 
allowance  for  waste  in  squaring  ? 

23.  A  man  wishes  to  lay  a  dressed-plank  walk  4  ft.  wide 
and  150  ft.  long  ;  the  planks  before  being  dressed  were 
1^  in.  by  9  in.  by  16  ft.  long.    How  many  board  feet  of 
lumber  must  he  buy,  if  he  puts  3  sleepers  2  in.  by  4  in. 
under  the  planks,  to  which  to  nail,  allowing  for  no  waste  ? 

24.  If  there  are  29  4-in.  spikes  in  a  pound,  how  many 
pounds  are  required  to  build  the  above  walk,  if  6  spikes 
are  used  in  each  piece  of  plank  ? 


10 


SHOP  PROBLEMS  IX  MATHEMATICS 


25.  Find  the  cost  of  laying  a  board  walk  on  the  side  and 
end  of  a  city  corner  lot  40  ft.  wide  and  120  ft.  long,  with 
the  following  specifications  : 

Width  of  walk,  4  ft. 

Boards  to  be  li  in.  by  7f  in.,  dressed  and  laid  cross- 
wise. 

Boards  to  be  placed  \  in.  apart. 

3  sleepers  placed  underneath,  3  in.  by  4  in. 

lOd.'wire  nails  to  be  used  (70  nails  in  a  pound,  at  5 
cents  per  pound),  and  6  nails  to  be  driven  into  each  board. 

The  lumber  to  cost  $35  per  M  and  the  labor  to  cost  $10. 


FIG.  6.    DRAWING  BOARD 


FIG.  7.    Box 


26.  A  drawing  board  23  in.  wide,  31  in.  long,  and  f  in. 
thick  is  constructed  of  5  pieces  joined  edge  to  edge  (see 
Fig.  6).    Allowing  J  in.  on  the  width   of   each   piece   for 
jointing  and  ^  in.  all  around  the  outer  edge  for  truing  up, 
how  many  board  feet  are  required  to  construct  the  board 
without  the  cleats  ? 

27.  A  covered  box  (Fig.  7)  is  constructed  of  1-in.  stock. 
Its  outside  dimensions  are  as  follows  : 

Height,  14  in. 

Width,   24  in. 

Length,  42  in. 

At  $70  per  M,  what  will  be  the  cost  of  lumber  for  the  box, 
allowing  1  sq.  ft.  for  waste  in  squaring  ? 


BOARD  MEASURE  11 

28.  How  many  board  feet  of  lumber  can  be  loaded  into 
a  car  36  ft.  long  and  8  ft.  wide,  if  the  lumber  is  f  in.  thick 
and  is  piled  6  ft.  high  ? 

29.  How  many  feet  of  lumber  will  be  required  for  every 
12  ft.  of  fencing  constructed  of  1-in.  stock,  the  fence  to  be 
7  ft.  high,  allowing  2  posts  4  in.  by  4  in.  by  10  ft.  long, 
and  2  scantlings  2  in.  by  4  in.  by  12  ft.  long  for  each  12  ft.  ? 

30.  What  width   and   length  of  surfaced  boards,  |  in. 
thick,  would  you  select  for  a  tight,  plain,  covered  box  3  ft. 
11  i  in.  long,  13i  in.  high,  and  23^  in.  wide,  outside  measure? 

The  edges  of  the  pieces  must  be  jointed  before  the  box  is  nailed 
together  ;  hence  \  in.  will  be  taken  from  their  width. 

31.  How  many  feet  are  required  for  the  box  mentioned 
in  Problem  30,  and  how  much  lumber  is  actually  wasted  in 
its  construction  ? 

32.  The  inside  dimensions  of  a  box  without  a  cover  are 
9|  in.  deep,  2  ft.  9£  in.  wide,  and  2  ft.  10  in.  long.    State, 
first,  what  size  of  boards  you  would  select ;  second,  how 
many  feet  they  would  contain;   third,  how  many  feet  of 
lumber  the  box  would  actually  contain  when  finished,  the 
lumber  to  be  surfaced  and  to  be  f  in.  thick. 

33.  A  man  wishes  to  build  a  picket  fence  on  the  side  of 
his  lot.   The  distance  between  the  two  ends  of  the  fence,  if 
measured  on  a  level,  is  exactly  150  ft.    Because  of  a  hill 
in  the  middle  of  the  lot  over  which  the  fence  must  pass  he 
finds  its  length  to  be  204  ft.    How  much  will  the  fence 
cost  according  to  the  following  specifications  ? 

Pickets  3  in.  wide  and  placed  3  in.  apart  (cost  5  cents 
each). 

Two  rails  on  which  to  nail  pickets,  2  by  4  in.  (cost 
$24  per  M). 

Two  posts  every  12  ft.  (at  25  cents  each). 


12  SHOP  PROBLEMS  IN  MATHEMATICS 

Each  picket  to  receive  4  nails  (8d.  wire  ;  95  nails  in  a 
pound,  ,at  5  cents  per  pound). 

The  rails  to  have  2  spikes  in  each  post  (20d.  wire  ;  28 
spikes  in  a  pound,  at  4  cents  per  pound). 

Labor  to  cost  $12. 

As  an  aid  to  solving  this  problem,  it  is  suggested  that  the  student 
draw  a  diagram  of  a  fence  running  over  a  hill,  and  also  of  one  on  the 
level  between  the  same  two  points,  and  compare  the  number  of  pickets. 
Pickets  are  placed  plumb. 

34.  A  load  of  lumber  consists  of  l|-in.  boards  12  in. 
wide  and  16  ft.  long,  placed  edge  to  edge.    How  many  feet 
does  the  load  contain,  if  it  is  36  in.  wide  and  30  in.  high  ? 
(Use  a  short  method.) 

35.  A  carload  of  lumber  consists   of   3  tiers   of  2*-in. 
planks,  the  tiers  being  12  ft.  long,  8  ft.  2  in.  wide,  and  8  ft. 
4  in.  high.   How  many  feet  does  the  car  contain,  if  we  deduct 

2  in.  from  the  width  of  each  layer  for  cracks  between  the 
edges  of  the  planks  ? 

36.  A  stack  of  1^-in.  boards  is  16  ft.  long,  12  ft.  wide, 
and  13  ft.  5  in.  high.    To  allow  for  seasoning,  each  layer  of 
boards  is  separated  from  the  one  below  by  3  pieces,  each 
1  in.  by  6  in.  by  12  ft.  long  (one  at  either  end,  and  one  in 
the  middle  running  crosswise  of  the  stack).    If  we  deduct 

3  in.  from  the  width  of  the  stack  for  cracks  between  the 
edges  of  the  boards,  and  count  in  the  crosspieces,  how 
many  feet  does  the  stack  contain  ? 

37.  How  many  feet  would  there  be  in  the  stack  of  lum- 
ber mentioned  in  Problem  36,  if  the  boards  were  1^  in. 
thick  and  the  cross  pieces  4  in.  instead  of  6  in.  wide  ? 

38.  A  class  in  manual  training  makes  36  ink  trays  (see 
Fig.  8)  out  of  ash,  each  tray  being  constructed  of  two  pieces, 
the  base  and  the  top.  When  finished  the  base  is  j|  in.  thick, 
5f  in.  wide,  and  8|  in.  long ;  and  the  top  T%  in.  thick,  3i  in. 


BOARD  MEASURE  13 

wide,  and  5|  in.  long.  Allowing  1  in.  on  the  length  and  J-  in. 
on  the  width  of  each  piece  for  rough  stock,  how  much  will 
lumber  for  the  trays  cost  at  $80  per  M  for  $-in.  stock  and 
$50  per  M  for  f-in.  stock  ? 

39.  Find  the  cost  for  material  for  72  tool  racks,  if  each 
rack  requires  the  following  : 

1  piece  2  in.  by  2  in.  by  15  in.  white  wood  at  $70  per  M. 

1  piece  |  in.  by  4  in.  by  15  in.  white  wood  at  $60  per  M. 

3  flat-head  screws  li  in.,  #  7,  at  10  cents  per  gross. 


FIG.  8.    PEN  AND  INK  TRAY 

40.  At  $60  per  M  for  white  wood,  $80  per  M  for  ash,  15 
cents  per  gross  for  1-in.  ^  6  round-head  screws,  and  10  cents 
per  gross  for  burrs  (washers),  what  will  the  material  for  100 
drawing  boards  cost,  the  boards  to  be  constructed  as  follows  ? 
Size  of  board,  14  in.  by  17  in.  by  1  in.  thick. 
To  be  constructed  of  4  pieces,  jointed  edge  to  edge. 
To  have  2ash  cleats  f  in.  thick,  1  ^  in.  wide,  and  13  in.  long. 
Each  cleat  to  be  fastened  to  board  with  5  round-head 
screws,  with  burr  under  head. 

Cost  of  glue  for  each  board,  2  cents. 
Allow  i  in.  on  the  width  of  each  of  the  4  pieces  in  the 
board  for  jointing,  and  1  in.  on  the  length  for  truing  up  ; 
also  I  in.  on  the  width  and  1  in.  on  the  length  of  the 
cleats  for  rough  stock. 

18.  A  formula.  A  formula  is  a  brief  statement  of  a  rule. 
For  example,  "  The  area  of  a  rectangle  is  equal  to  the  base 
multiplied  by  the  altitude  "  may  be  stated  R  =  b  x  a. 


14 


SHOP  PROBLEMS  IN  MATHEMATICS 


EXERCISE 

If  b  is  the  number  of  board  feet,  t  the  number  of  inches 
thick,  w  the  number  of  inches  wide,  and  I  the  number  of 
feet  long,  write  the  formula  for  b  in  terms  of  £,  w,  and  L 
State  the  rule  which  this  formula  represents.  Using  the 
formula,  find  b  in  pieces  of  lumber  having  the  following 
dimensions  : 

1.  61  in.  by  8  in.  by  14  ft. 

2.  4  in.  by  10  in.  by  12  ft. 

3.  2  in.  by  8  in.  by  16  ft. 

4.  3  in.  by  9  in.  by  14  ft. 


5.  2£  in.  by  9  in.  by  10  ft. 

6.  i  in.  by  9  in.  by  16  ft. 

7.  |  in.  by  7  in.  by  10  ft. 

8.  I  in.  by  10  in.  by  14  ft. 
9.  £  in.  by  9  in.  wide  at  one  end,  15  in.  wide  at  the  other, 

and  12  ft.  long.    (Take  w  equal  to  the  average  width.) 

19.  Method  of  com- 
puting surface  measure 
approximately.  A  handy 
method  to  use  in  calcu- 
lating the  square  feet 
in  a  rectangular  sur- 
face, where  the  dimen- 
sions are  in  feet  and 
inches,  is  shown  in  the 
following : 

Take,  for  example, 
a  rectangle  3  ft.  7  in. 
by  4  ft.  6  in.  (Fig.  9). 
If  we  begin  at  one 
corner,  lay  off  feet  and 

draw  lines  as  shown  in 
FIG.  9.   APPROXIMATE  SURFACE  MEASURE    the   figure?   we    get   12 

sq.  ft.,  represented  by  the  letter  a,  with  7  in.  over  on  one 
side  and  6  in.  over  on  the  other. 


a 

a 

a 

b 

£ 
<T 

a 

a 

a 

C 

Q 

Q 

a 

d 

a 

a 

a 

e 

f 

9 

h 

i 

'1 

e                    •>  ft             -^ 

^-7^. 

p                                Orf- 

^  /^ 

BOARD  MEASURE  15 

ACTUAL  WORK  EXPLANATION 

3  ft.     7  in.     4  ft.  x  3  ft.  =  12  sq.  ft.  (Area  indicated  by  a) 

4  6          4  ft.  x  7  in.  =  f  8  sq.  ft. 

12  =  2sq.ft.andT\sq.ft.  (b,c,d,e) 

2  4          6  in.  x  3  ft.  =  ||  sq.  ft. 

1  6  =  1  sq.  ft.  and  ^  sq.ft.  (f,g,h) 

3          6  in.  x7in.  =  42  sq.  in. 

16  2  =y\sq.  ft.  and  T|4 :  sq.  ft.  (i) 

The  student  will  observe  that  the  first  column  contains 
square  feet  and  the  second  twelfths  of  a  square  foot.  A 
remainder  equal  to  6  sq.  in.  (T|¥  sq.  ft.)  or  more  is  counted 
as  ^  sq.  ft.  and  added  to  the  second  column.  Thus,  in  the 
example,  T£4-  sq.  ft.  is  counted  as  T^  sq.  ft.  and  added  to 
the  second  column,  giving  a  sum  of  {f  sq.  ft.,  which  is 
equal  to  1  sq.  ft.  and  ^  S(l-  ^.  Adding  1  sq.  ft.  to  the  first 
column,  we  have  16  sq.  ft.  and  ^  sq.  ft.  or  16^  sq.  ft. 

This  method  is  accepted  by  the  United  States  government  as  a 
standard  method  on  all  such  material  as  imported  granite  and  marble. 

EXERCISE 

By  the  above  method  find  the  number  of  square  feet  and 
fractions  of  a  square  foot  in  surfaces  having  the  following 
dimensions  : 

1.  1  ft.  by  1  ft.  6  in.  10.  12  ft.  11  in.  by  10  ft.  5  in. 

2.  1  ft.  by  1  ft.  9  in.  11.  16  ft.  4  in.  by  12  ft.  7  in. 

3.  2  ft.  by  2  ft.  6  in.  12.  17  ft.  8  in.  by  9  ft.  11  in. 

4.  4  ft.  by  3  ft.  4  in.  13.  10 £  ft.  by  9£  ft. 

5.  5  ft.  by  3  ft.  9  in.  14.  lift  ft.  by  7}i  ft. 

6.  2  ft.  by  11  in.  15.  12J,  ft.  by  16  ft.  9  in. 

7.  3  ft.  6  in.  by  1  ft.  6  in.  16.  16f  ft.  by  16|  ft. 

8.  5  ft.  6  in.  by  3  ft.  7  in.  17.  18  ft.  10  in.  by  13f  ft. 

9.  10  ft.  8  in.  by  6  ft.  5  in.  18.  20  ft.  6  in.  by  16J  ft. 


16  SHOP  PROBLEMS  IN  MATHEMATICS 

19.  17  ft.  7  in.  by  17^  ft-      22.  40  ft.  7  in.  by  26§  ft. 

20.  30  ft.  7  in.  by  20  ft.  9  in.     23.  29  ft.  4  in.  by  20  ft.  7  in. 

21.  25  ft.  4  in.  by  35  ft.  3  in.     24.  33£  ft.  by  15  ft.  4  in. 

PROBLEMS 

1.  How  many  feet  of  lumber  will  be  required  to  floor  a 
platform  16  ft.  4  in.  long  by  11  ft.  6  in.  wide,  the  lumber 
to  be  1  in.  thick,  if  no  allowance  is  made  for  waste  ? 

2.  The  foundation  of  a  house  is  22  ft.  4  in.  by  40  ft.  6  in. 
How  much  |-in.  surfaced  lumber  will  it  take  for  the  lower 
subfloor,  if  we  cut  out  a  place  for  the  cellar  stairs  3  ft. 
3  in.  by  7  ft.  7  in.,  making  no  allowance  for  waste  ? 

3.  The  distance  from  the  bottom  of  the  sill  to  the  top  of 
the  plate  on  a  cottage  house  is  16  ft.  10  in.    If  the  house  is 
25  ft.  6  in.  long  and  we  deduct  £  for  openings  (windows 
and  doors),  what  will  be  the  cost  of  sheathing  one  side  with 
£-in.  surfaced  lumber  at  $35  per  M  ?    (See  Fig.  14.) 

4.  At  15/  per  square  foot,  what  will  be  the  cost  of  laying 
a  cement  sidewalk  4  ft.  11  in.  wide  and  50  ft.  8  in.  long  ? 

5.  At  $36  per  M,  how  much  will  hemlock  lumber  cost 
for  the  following  platform  ? 

Size  of  platform  16  ft.  11  in.  long  and  14  ft.  wide. 
Supports  3  in.  by  6  in.  by  14  ft.,  placed  2  ft.  5  in.  from 
center  to  center,  running  crosswise  (rough). 

Floor  to  be  of  IJ-in.  by  9|-in.  boards  (dressed). 

6.  A  drawing  board  is  1  ft.  11  in.  wide  and  2  ft.  7  in.  long. 
What  is  the  size  of  its  surface  in  feet  ? 


CHAPTER  II 

HOUSE  BUILDING,  GENERAL  CONSTRUCTION,  HEIGHTS  OF 
TREES  AND  OTHER  MEASUREMENTS 

FLOORING 

20.  Flooring.  The  term  "  flooring  "  is  sometimes  used  to 
designate  the  floor  of  a  building  after  it  is  laid;  but  for 
brevity  we  shall  use  the  term,  unless  otherwise  stated,  to 
mean  common  matched  flooring  lumber  before  it  is  laid. 


FIG.  10.    FLOORING 

21.  Allowance  for  matching  and  waste.  When  estimating 
flooring,  ceiling,  or  other  lumber  that  is  matched,  i.e.  has  a 
tongue-and-groove  joint  (see  Fig.  10),  enough  stock  must  be 
added  to  make  up  for  the  amount  cut  away  from  the  width 
in  matching.  This  amount  varies  from  J  in.  to  f  in.  on  each 
board,  according  to  the  size  and  nature  of  the  stock  matched. 

Because  of  the  uneven  nature  of  lumber  and  different 
methods  of  working  it,  it  is  impossible  to  formulate  any 

17 


Of   THt 

UNIVERSITY 


18  SHOP  PROBLEMS  IN  MATHEMATICS 

exact  mathematical  rule  showing  the  relation  of  the  feet  in 
a  lot  of  lumber  before  and  after  it  is  matched.  However,  it 
can  be  estimated  approximately,  and  near  enough  for  all 
practical  purposes. 

A  little  flooring  is  always  wasted  in  squaring  the  ends, 
cutting  up,  etc.  To  offset  this  the  mechanic  usually  adds 
a  few  feet  to  his  bill,  enough,  in  his  estimation  from  ex- 
perience, to  make  up.  For  the  sake  of  convenience,  as  well 
as  for  the  sake  of  fixing  the  subject  in  the  student's  mind, 
an  allowance  for  this  purpose  will  be  added  in  the  follow- 
ing problems.  The  amount  allowed  varies  from  one  to  ten 
per  cent,  according  to  the  size  and  nature  of  the  flooring. 

For  flooring,  ceiling,  etc.,  from  about  21  in.  up  to  5i  in. 
in  width,  the  amount  added  for  matching  is  generally 
one  fourth.  For  the  so-called  "  Adams  "  flooring  or  pieces, 
usually  about  1^  in.  wide,  the  amount  added  is  one  third. 
For  example,  if  a  common  floor  to  be  laid  is  12  ft.  square, 
the  amount  of  flooring  required  is  144  ft.  plus  £  of  144,  or 
36  ft.,  making  in  all  180  ft.  If  the  floor  to  be  laid  is  of 
pieces  1^  in.  wide,  the  amount  required  is  144  ft.  plus 
48  ft.,  or  192  ft. 

22.  Face.  In  most  localities  it  is  customary  to  give  the 
lumber  dealer  the  actual  size  of  the  surface  to  be  covered, 
called  "  face,"  and  let  him  add  the  proper  amount  for 
matching. 

PROBLEMS 

1.  How  much  |-in.  flooring  is  required  to  lay  a  floor  16  ft. 
by  18  ft.,  if  we  add  3  per  cent  for  squaring  the  pieces  ? 

2.  How  much  hard-pine  flooring  |  in.  thick  is  required 
for  a  floor  13  ft.  6  in.  by  14  ft.  10  in.,  4  per  cent  being 
allowed  for  waste  ? 

3.  The  foundation  of  a  house  is  26  ft.  by  30  ft.    At  $60 
per  M,  what  will  be  the  cost  of  flooring  for  one  floor,  if 


HOUSE  BUILDING  19 

the  flooring  extends  within  5  in.  of  every  outside  edge,  on 
account  of  the  studding,  and  there  is  an  opening  for  cellar 
stairs  3  ft.  4  in. by  8  ft.  2  in.?  Allow  3  per  cent  for  squaring. 
(See  Fig.  14.) 

4.  A  kitchen  floor  is  10  ft.  6  in.  at  one  end,  9  ft.  2  in. 
at  the  other,  and  11  ft.  7  in.  long.    If  it  is  laid  with  hard- 
maple  flooring  1|  in.  wide,  what  will  the  lumber  cost  at 
$60  per  M?    Add  4  sq.  ft.  for  waste. 

5.  The  floor  of  a  veranda  is  laid  of  stock  li  in.  thick. 
If  the  size  of  the  floor  is  7  ft.  11  in.  wide  and  22  ft.  6  in. 
long,  how  many  feet  of  flooring  does  it  require,  allowing 
2  per  cent  for  squaring  ?    (See  sect.  5.) 

6.  The  veranda  on  a  corner  of  a  house  runs  18  ft.  6  in. 
across  the  front  and  13  ft.  4  in.  down  the  side.    If  it  is  7  ft. 
6  in.  wide  and  we  use  l|-in.  flooring,  how  many  feet  must 
we  buy,  2  per  cent  being  added  for  squaring  ? 

7.  How  many  squares  (100  sq.  ft.)  of  floor  are  there  in 
a  four-story  house  27  ft.  wide  by  41  ft.  4  in.  long  within 
walls,  deducting  from  every  floor  an  opening  10  ft.  by 
4  ft.  4  in.  for  stairs;  and  how  many  feet  of  common  hard- 
pine  flooring  does  it  require,  adding  2  per  cent  for  squar- 
ing, if  the  flooring  is  £  in.  thick  ? 

8.  The  floor  of  a  dancing  pavilion  is  octagonal  in  form. 
If  it  is  12  ft.  on  a  side,  how  much  hard-pine  flooring  |  in. 
thick  and  2^  in.  wide  is  required,  allowing  3  per  cent  for 
waste  in  squaring? 

9.  A  dealer's  bill  for  flooring  was  $200.   If  the  price  was 
$45  per  M,  state,  first,  how  many  feet  of  lumber  (flooring) 
there  were;   second,  how  many  square  feet  of  floor  was 
covered ;  and  third,  what  was  the  length  of  one  side  of 
the  octagonal  floor  laid  ?    No  waste. 

10.  At  $45  per  M,  what  will  be  the  cost  of  flooring  li  in. 
wide  for  a  room  13  ft.  9  in.  wide  and  16  ft.  7  in.  long,  the 


20  SHOP  PROBLEMS  IN  MATHEMATICS 

room  to  have  a  bay  window  with  octagonal  corners  at  one 
end  2  ft.  deep  and  6  ft.  long  across  the  front,  allowing  5  per 
cent  for  waste  in  cutting  ? 

11.  How  many  feet  of  flooring  are  required  for  one  floor 
of  a  circular  tower,  the  radius  of  which  is  7  ft.  4  in.  within 
walls  ?    Add  5  per  cent  for  squaring. 

12.  A  room  14  ft.  7  in.  wide  and  15  ft.  8  in.  long  has  a 
circular  window  extending  from  one  corner.    The  radius  of 
the  window  is  4  ft.  10  in.  within  walls,  and  its  center  is  at 
the  point  where  the  corner  of  the  room  would  be  if  the 
window  were  not  there.    Allowing  5  per  cent  for  waste  in 
squaring  up  the  pieces,  how  many  feet  of  flooring  1^  in. 
wide  are  required  for  the  room  ? 

13.  With  ^-in.  ceiling,  at  $30  per  M,  how  much  will  the 
lumber  cost  to  ceil  the  four  walls  and  overhead  of  a  room 
12  ft.  6  in.  by  18  ft.  8  in.  by  10  ft.  2  in.  high,  if  there  are 
three  openings  for  doors  3  ft.  4  in.  by  7  ft.  6  in.  each,  and 
2  openings  for  windows  3  ft.  7  in.  by  7  ft.  2  in.  each,  the 
ceiling  to  be  2£  in.  and  over  wide,  and  25  ft.  to  be  added 
for  waste  in  squaring  ? 

14.  The  dining  room  of  a  house  is  to  be  floored  with 
white-oak  flooring  f  in.  thick  and  1^  in.  wide.    It  is  also  to 
be  wainscoted  3  ft.  up  from  the  floor  with  3-in.  white  oak 
matched  stock  J  in.  thick.    If  we  deduct  for  2  doors  3  ft.  4  in. 
each,  and  for  2  windows  running  within  18  in.  of  the  floor, 
each  3  ft.  6  in.  wide ;  if  the  room  is  11  ft.  5  in.  wide  and 
16  ft.  3  in.  long,  and  we  add  5  per  cent  for  waste,  how 
many  feet  of  flooring  and  wainscoting  are  required  ? 

15.  The  sides  of  an  automobile  garage  are  covered  with 
3£-in.  white  pine  beaded  ceiling  -J  in.  thick.    The  size  of 
the  house  on  the  ground  is  12  ft.  by  16  ft.  9  in.,  the  roof 
is  a  common  gable  (see  Fig.  15),  the  peak  rises  above  the 
level  of  the  plates  4  ft.,  and  the  corner  posts  are  9  ft.  long. 


HOUSE  BUILDING 


21 


If  we  take  out  2  openings  2  ft.  10  in.  by  4  ft.  5  in.  each  for 
windows,  and  one  opening  8  ft.  wide  and  8  ft.  high  for  a 
door,  how  many  feet  of  ceiling  are  required  ?  Add  16  sq.  ft. 
for  waste. 

SIMPLE  STAIR  BUILDING 

23.  Rise.  The  "  rise  "  of  a  flight  of  stairs  is  the  perpen- 
dicular distance  the  stairs  are  to  rise,  e.g.  the  distance  be- 
tween any  two  floor  levels  of  a  building  (see  Fig.  11).  In 


QJ 

-9 

or 


Upper  Flo  or  L  in  e 


Width  of  Riser 

XT 
Run 


r 


Lower  Floor  Line 


1- Run  •* 

FIG.  11.    STAIRS 

other  words,  it  is  the  sum  total  of  the  widths  of  all  the 
risers  (see  Fig.  12). 

24.  Nosing.    The  treads  of  a  flight  of  stairs  almost  always 
project  over  the  front  of  the  risers.   This  projection  is  called 
"  nosing."    Its  amount  is  about  the  same  as  the  thickness  of 
the  tread  (see  Fig.  12). 

25.  Run.    The  "run"  of  a  flight  of  stairs  is  the  distance, 
measured  on  a  level,  from  the  foot  of  the  stairs  to  a  point 
directly  under  their  upper  end  (see  Fig.  11).    It  is  the  hori- 
zontal space  the  stairs  take  up,  and  is  the  sum  of  the  widths 
of  the  treads  minus  their  nosing  plus  the  width  of  one  tread. 


22  SHOP  PROBLEMS  IN  MATHEMATICS 

26.  Number  of  risers  and  treads.    There  is  always  one  more 
riser  than  treads,  one  floor  counting  as  a  tread. 

27.  Number  of  steps.    To  get  the  desired  number  of  steps 
for  any  common  flight  of  stairs,  divide  the  rise  of  the  stairs 
by  the  even  inch  nearest  the  desired  height  of  the  steps. 
This  generally  gives  a  whole  number  and -a  fraction.    Now, 
in  turn,  divide  the  rise  again  by  the  whole  number  obtained 
above,  disregarding  the  fraction,  and  the  result  will  be  the 
required  width  of  each  riser.    (The  whole  number  obtained 
above  will  be  the  number  of  steps.)    For  example,  the  rise 
of  a  flight  of  stairs  is  9  ft.  8  in.,  and  it  is  desired  to  have 
the  steps  as  near  7  in.  high  as  possible.    9  ft.  8  in.  reduced 
to  inches  and  divided  by  7  equals  16f  ;  now  if  we,  in  turn, 
divide  the  9  ft.  8  in.  by  16,  we  have  7|  in.,  the  width  of 
each  riser  or  the  height  of  each  step. 

The  width  of  the  treads  is  obtained  by  dividing  the  run 
of  the  stairs  by  the  number  of  steps  or  risers  and  then 
adding  to  this  the  amount  allowed  for  nosing. 

PROBLEMS 

1.  The  risers  (Fig.  11)  of  a  flight  of  stairs  are  7  in.  wide. 
How  many  steps  will  there  be,  if  the  rise  of  the  stairs  is 
3ft.  6  in.? 

2.  The  height  of  the  first  floor  of  a  house  above  the  cellar 
bottom  is  8  ft.  2  in.,  and  the  run  of  the  cellar  stairs  is  8  ft. 
If  there  are  13  steps,  what  will  be  the  width  of  both  the 
treads  and  the  risers,  allowing  1  in.  for  nosing  ? 

3.  If  the  height  of  a  veranda  floor  is  31^  in.  from  the 
ground,  and  we  make  the  risers  as  nearly  as  possible  7  in., 
how  many  risers  are  required,  and  what  will  be  their  width  ? 

4.  What  is  the  length  of  a  stair  string,  if  the  rise  is  8  ft. 
10  in.  and  the  run  is  8  ft.  2  in.,  making  no  allowance  for 
waste  in  cutting  ?    (See  Fig.  12.) 


HOUSE  BUILDINJ& 


23 


5.  The  rise  of  a  flight  of  stairs  is  10  ft.  6  in.  and  the 
run  is  11  ft.    How  many  feet  of  lumber  are  required  for 
the  two  strings,  if  they  are  |  in.  thick,  10  in.  wide,  and 
we  allow  IL  ft.  on  the  length  of  each  string  for  cutting 
on  a  slant  ? 

6.  The  distance  between  the  first  and  second  floors  of  a 
house  is  12  ft.  4  in.  and  the  run  of  the  stairs  is  13  ft.  8  in. 


FIG.  12.    STAIRS 

If  we  make  the  risers  as  near  7  in.  as  possible,  what  will 
be  the  number  of  treads  and  their  width,  if  they  project  over 
the  risers  1J  in.,  and  what  will  be  the  number  and  width  of 
the  risers  ? 

7.  How  many  feet  of  lumber  are  required  for  the  steps 
(treads  and  risers)  of  a  front  stoop  the  height  of  which  is 
4  ft.  1  in.,  the  run  4  ft.  4  in.,  and  the  length  7  ft.  10  in., 
the  risers  to  be  of  J-in.  stock,  and  the  treads  to  be  of  IJ-in. 
stock  and  to  project  over  the  risers  1^  in.?  Allow  1  in.  on 


24  SHOP  PROBLEMS  IN  MATHEMATICS 

either  end  of  every  piece  for  squaring,  and  J  in.  on  the 
width  for  jointing  up. 

8.  With  oak  lumber  at  $80  per  M,  what  will  the  lumber 
cost  for  the  following  flight  of  stairs  ? 

Eise,  13  ft. 

Eun,  14  ft.  7  in. 

Steps  to  be  as  near  7  in.  as  possible. 

Steps  to  be  3  ft.  11  in.  long. 

Strings  to  be  -J  in.  thick  by  11  in.  wide. 

Treads  to  be  1 J  in.  thick  and  to  project  over  risers  1 J  in. 

Eisers  to  be  -J  in.  thick. 

Allow  1  in.  on  the  length  of  the  treads  and  risers  for 
squaring,  J-  in.  on  the  width  of  all  pieces  for  jointing,  and 
1^-  ft.  on  the  length  of  each  string  for  cutting  on  a  slant. 

9.  The  distance  between  floors  in  a  house  is  11  ft.  8  in. 
The  conditions  are  such  that  the  stairs  must  go  part  way  to 
a  platform  and  then  turn  and  go  up  at  right  angles  the  rest 
of  the  way.    The  height  of  the  platform  above  the  first  floor 
is  4  ft.  4^-  in.    How  many  steps  does  each  flight  require  (the 
riser  to  be  as  near  7  in.  as  possible),  what  is  the  width 
of  the  risers  in  each  flight,  and  what  is  the  run  of  each 
flight,  if  the  treads  are  lOf  in.  wide  and  project  over  the 
risers  1J  in.  ? 

SIMPLE  FRAMEWORK 

28.  Ordering  rougher  lumber.  In  ordering  rougher  lumber 
only  standard  lengths  are  generally  considered.  This  some- 
times involves  the  purchase  of  a  few  more  feet  than  are 
actually  required.  The  expense,  however,  of  having  the 
lumber  cut  to  exact  length  would  be  greater  than  buying 
a  few  extra  feet  on  the  length  of  some  pieces  that  are  too 
long.  Take,  for  example,  floor  joists  that  are  13  ft.  6  in. 
when  placed  ;  the  nearest  standard  length  is  14  ft.  It  would 


HOUSE  BUILDING 


25 


be  cheaper  to  buy  the  14-ft.  lengths  and  throw  away  the 
extra  6  in.  than  to  order  the  pieces  cut  13  ft.  6  in.  In  either 
case  the  same  amount  of  lumber  would  probably  be  charged 
for,  besides  the  extra  work  in  cutting  off  the  pieces. 

When  practicable,  long  pieces  can  be  obtained  and  cut  in 
two  for  short  ones.  This,  however,  can  be  done  with  profit 
only  up  to  a  certain  limit,  as  all  lengths  above  16  ft. 


FIG.  13.    HOUSE  FOUNDATION 

increase  in  price  in  proportion  as  the  length  increases.  In 
some  localities  the  price  increases  with  the  length  above 
12ft. 

PROBLEMS 

1.  How  many  feet  are  required  for  the  sills  (see  Fig.  13) 
of  a  building  20  ft.  wide  by  40  ft.  long,  the  timber  to  be 
6  by  8  in.  by  16  ft.  long,  if  we  deduct  1  ft.  from  the  length 
of  each  16-ft.  piece  for  splices  ? 

2.  The  floor  foundation  (sills  and  floor  joists)  of  a  build- 
ing is  16  ft.  6  in.  by  24  ft.,  the  sills  are  6  by  8  in.,  the  floor 
joists  (see  Fig.  13)  are  2  by  10  in.,  and  are  placed  across  the 


26 


SHOP  PROBLEMS  IN  MATHEMATICS 


narrow  way  of  the  building,  16  in.  from  center  to  center. 
What  standard  lengths  of  lumber  would  you  select,  and 
how  many  feet  are  required  for  the  foundation,  if  splices 
on  the  sills  take  up  1  ft.  on  every  piece,  no  piece  to  be  over 
16  ft.  long  ? 

3.  With  framing  lumber  at  $30  per  M,  how  much  will 
the  lumber  cost  for  the  sills  and  lower  floor  joists  of  a 


FIG.  14.    FRAME  OP  HOUSE 

house,  with  the  following  dimensions  ?  State  also  the 
lengths  of  the  lumber  you  would  select,  nothing  to  be 
over  16  ft.  long. 

Size  of  house  to  outside  of  sills,  25  ft.  by  44  ft.  6  in. 

Size  of  sills,  6  by  8  in. 

Two  rows  of  floor  joists  placed  16  in.  from  center  to 
center,  running  across  the  narrow  way  of  the  building, 


HOUSE  BUILDING  27 

with  one  end  resting  on  a  girder  in  the  center  of  the 
cellar. 

Splices  to  be  1  ft.  long  on  sills. 

The  studding  (see  Fig.  14)  and  floor  joists  in  common  house  work 
are  placed  16  in.  from  center  to  center  so  that  standard  laths  (4  ft. 
long)  can  be  used  without  waste. 

4.  The  partition  in  a  house  is  8  ft.  high  and  16  ft.  long. 
If  it  is  studded  with  2-  by  4-in.  pieces  placed  16  in.  from 
center  to  center,  how  many  pieces  are  required  ?    How 
many  feet  do  these  pieces  contain,   and  what  standard- 
length  stock  would  you  select  ? 

5.  If  the  studding  are  3  by  4  in.  by  18  ft.,  and  are  placed 
16  in.  from  center  to  center,  how  many  are  required  for  one 
side  of  a  house  36  ft.  long;  and  how  many  feet  do  they 
contain,  allowing  10  extra  studding  for  double  corners  and 
double  studs  at  the  windows  ? 

6.  At  $30  per  M,  what  will  be  the  cost  of  frame  lumber 
for  the  walls  and  floors  of  the  following  building,  and  how 
many  pieces  of  each  kind  should  there  be? 

Size  of  building  on  the  ground,  16  ft.  by  20  ft. 

Height  of  building  from  bottom  of  sills  to  top  of 
plates  (see- Fig.  14),  16  ft.  8  in. 

Sills  to  be  6  in.  square  and  no  longer  than  16  ft.  for 
each  piece. 

Lower  floor  joists  to  be  2  by  8  in.,  placed  16  in.  from 
centers. 

Joists  for  2  upper  floors,  2  by  6  in.,  placed  16  in.  from 
centers. 

Studding  for  4  sides,  2  by  4  in.,  with  16  extra  for 
double  corners  and  windows. 

Plates  to  be  2  by  4  in.,  doubled. 

Add  twenty  £  by  6  in.  by  12  ft.  boards  for  stays, 
braces,  etc. 


28 


SHOP  PROBLEMS  IN  MATHEMATICS 


ROOF  WORK 

29.  Length  of  rafters.  In  common  roof  construction  the 
lengths  and  various  cuts  of  rafters  are  usually  obtained  on 
the  steel  square  without  the  aid  of  much  mathematical 
calculation.  However,  knowledge  of  the  mathematical 
principles  involved  and  of  their  application  is  of  in- 
estimable value  to  the  mechanic. 


FIG.  15.    COMMON  GABLE  ROOF 

30.  Rise.    The  "  rise  "  of  a  rafter  is  the  height  of  the 
ridge  of  the  roof  above  the  level  of  the  plates. 

If  the  rafters  are  raised  and  project  over  the  plate,  as  shown  in 
Fig.  15,  the  rise  is  the  distance  indicated  in  the  figure. 

31.  Run.    The  "  run "  of  a  rafter  is  the  distance  from 
where  the  rafter  intersects  the  outer  edge  of  the  plate  to  a 
point  on  a  level  with  the  plates,  directly  under  the  upper 
end  of  the  rafter  (see  Fig.  15).    It  is  usually  one  half  the 
width  of  the  building. 

32.  Pitch.    The  term  "pitch"  means  the  slant  or  slope 
of  a  roof  (see  Fig.  16).-  The  terms  "  1  pitch,"  "  J  pitch," 
"  \  pitch,"   etc.,  mean  that  the  height  of  the  ridge  of  a 
roof  is  equal  to  1,  ^,   |,  etc.,  the  width  of  the  building 
or  span  of  the  roof.     For  example,  let  the  width  of  the 


HOUSE  BUILDING 


29 


building  be  18  ft.;   if  the  pitch  of  the  roof  is 
^,  the  rise  will  be  6  ft. 

In  order  to  figure  out  rafter  problems  in  roof  con- 
struction it  is  absolutely  necessary  that  the  student 
understand  the  principles  underlying  the  solution  of 
right-angled  triangles  (for  which  see  sect.  44-46). 

PROBLEMS 

1.  The  rise  of  a  lean-to  roof  (see  Fig.  17)  is 
4  ft.  and  its  run  is   12  ft.    What 

length  of  the  rafters  ? 

2.  What   length   of   rafters   is 
quired,   and   what  is   the   pitch 
common  gable  roof  (see  Fig.  15), 
if  the  width  of  the  building  is 
24  ft.  and  the  rise  of  the  roof  8 
ft.,  allowing  the  rafters  to  pro- 
ject 18  in.  over  the 

side  of  the  build- 
ing at  the  eaves  ? 

3.  If  the  rafters 
are  2  by  6  in.,  how 

many  feet  of  lumber  will  they  contain  for  a  lean-to  roof 
with  a  13-ft.  run  and  a  6-ft.  rise,  the  width  of  the  lean-to 

being  14  ft.,  and 
the  rafters  being 
placed  2  ft.  from 
center  to  cen- 
ter and  project- 
ing over  at  the 
eaves  18  in.  ? 
(See  sect.  28.) 

4.  The  width  of 
FIG.  17.   LEAN-TO  ROOF  a  house  is  22  ft. 


MIL 


I    I    I    I    I     I    I    I 


15  14  13  IZ  II    10    9    8    7 

Run 


6    5"  4    3 


FIG.  16.    PITCH 


tt 

£j  (f) 

=£ 

2 
en 
oO 


30 


SHOP  PROBLEMS  IN  MATHEMATICS 


and  its  length  is  34  ft.    If  the  rise  of  the  roof  is  10  ft.,  and 
the  rafters  are  2  by  5  in.,  placed  2  ft.  from  center  to  center, 


FIG.  18.    HIP  ROOF 

and  project  over  at  the  eaves  8  in.,  how  many  feet  are  re- 
quired for  the  rafters  of  the  roof  ? 

5.  The  rise  of  a  hip  rafter  (see  Fig.  18)  is  10  ft.  and  its 
run  is  15£  ft.    What  is  its  length  ? 

-  Run    — H 


6.  A  building  is  28  ft.  square  and  the  rise  of  its  roof  is 
11  ft.    What  is  the  length  of  its  hip  rafters  ? 


HOUSE  BUILDING  31 

7.  How  many  feet  are  there  in  4  hip  rafters,  2  by  8  in., 
if  the  building  is  20  ft.  square  and  the  rise  of  the  roof  is 
8  ft.,  allowing  20  in.  for  eaves  projection  ? 

8.  The  width  of  a  straight-gable  house  is  25  ft.    What  is 
the  length  of  the  rafters  if  the  roof  has  ^  pitch  and  the 
eaves  projection  is  18  in.  ?    How  many  feet  do  the  rafters 
contain  if  they  are  2  by  5  in.  ? 

9.  How  many  feet  of  lumber  are  required  for  the  rafters 
of  a  roof  that  has  £  pitch  and  a  span  of  31  ft.,  the  length 


Valley 
Rafter 

FIG.  20.   VALLEY  ROOF 

of  the  building  being  48  ft.,  the  rafters  2  by  5  in.,  placed 
2  ft.  from  center  to  center,  and  having  an  eaves  projec- 
tion of  22  in.  ? 

10.  A  gambrel-roof  house  (see  Fig.  19)  is  24  ft.  wide, 
the  first  or  bottom  set  of  rafters  has  §   pitch,   and  the 
top  set  ^  pitch.    If  the  break  or  joint  in  the  roof  comes 
7  ft.  in  from  the  side  of  the  building,  how  long  is  each 
set  of  rafters  ? 

11.  A  house  is  built  in  the  form  of  an  L,  the  width  of 
each  part  being  20  ft.    What  is  the  length  of  the  required 
valley  rafters  if  the  roof  has  £  pitch  ?    (See  Fig.  20.) 


32  SHOP  PROBLEMS  IN  MATHEMATICS 

12.  In  the  above  problem  what  will  be  the  length  of  the 
jack  rafters  (see  Fig.  20)  if  we  begin  at  the  top  and  place 
them  2  ft.  from  center  to  center  ? 

In  getting  the  length  of  a  jack  rafter  of  a  hip  roof,  the  thickness  of 
the  hip  rafter  can  generally  be  disregarded  if  the  measurement  is  taken 
between  the  extreme  points  of  the  jack  rafter. 

13.  An  octagonal  tower  is  4  ft.  on  a  side.    How  long  must 
the  corner  or  hip  rafters  be  if  the  rise  of  the  roof  is  10  ft.  ? 

14.  How  many  feet  are  necessary  for  the  hip  rafters  of 
the  roof  of  a  hexagonal  tower  5  ft.  on  a  side  and  having  a 
perpendicular  height  of  11  ft.,  the  rafters  to  be  2  by  4  in. 
and  to  project  over  at  the  eaves  20  in.? 

15.  A  tower  is  circular  in  shape.    Its  diameter  is  14  ft. 
and  its  roof  has  §  pitch.    If  the  rafters  are  2  by  5  in.  and 
we  place  them  30  in.  apart  on  the  plate,  how  much  lumber 
do  they  contain  ? 

16.  The  width  of  a  house  is  28  ft.  and  the  rise  of  the  roof 
is  12  ft.    What  are  the  lengths  of  the  jack  rafters  for  the 
hip,  if  we  place  them  2  ft.  from  center  to  center,  beginning 
at  the  bottom  end  of  the  hip  rafter  and  letting  them  project 
18  in.  at  the  eaves  ? 

17.  A  house  is  built  in  the  form  of  the  letter  T.  The  large 
or  top  part  of  the  T  is  22  ft.  wide  and  has  a  \  pitch  roof  ; 
the  smaller  part  is  18  ft.  wide  and  has  a  roof  that  rises  8  ft. 
What  is  the  length  of  the  common,  valley,  and  jack  rafters, 
the  jack  rafters  to  begin  at  the  bottom  end  of  the  valleys  and 
to  be  placed  2  ft.  from  center  to  center  ?  (See  note  below.) 

In  actual  practice  enough  must  be  added  to  the  length 
of  jack  rafters  for  valleys  to  allow  for  the  slant,  on  the 
lower  end  of  the  jacks,  that  fits  the  valley  rafter. 

The  actual  measurement  for  valley  jacks  is  generally  taken  from 
the  lower  short  corner  on  the  lower  end  to  the  extreme  point  on  the 
upper  end,  the  thickness  of  the  valley  being  disregarded. 


HOUSE  BUILDING  33 

SHINGLING 

33.  Standard  shingles  are  figured  on  a  basis  of  16  in.  long 
and  4  in.  wide,  and  are  estimated  by  the  thousand.    They 
are  put  up  in  bunches  of  250  each.;  hence  4  bunches  to 
the  thousand. 

The  16-in.  shingle,  when  laid,  is  usually  exposed  to  the 
weather  about  5  in.  (called  "  5  in.  to  the  weather  "). 

34.  Number  of  shingles.    There  are  various  methods  of 
calculating  the  number  of  shingles  for  a  roof,  the  most 
common  of  which  are  the  following  : 

(1)  Find  the  number  of  squares  (100  sq.  ft.)  in  the  roof ; 
divide  this  number  by  1^  and  multiply  by  1000. 

EXAMPLE.    A  roof  is  25  ft.  long  and  20  ft.  wide. 
25  x  20  =  5  squares.    —  x  1000  =  4000,  the   number    of 
shingles  required. 

(2)  Multiply  the  number  of  square  feet  in  the  roof  by  1\. 

EXAMPLE.  The  length  of  the  ridge  of  a  roof  is  32  ft.  and 
the  length  of  the  rafters  on  either  side  is  18  ft. 

32  x  2  x  18  x  7  J  =  8640,  the  number  of  shingles  required. 

(3)  As  it  takes  approximately  3  bunches  (250  shingles 
each)  to  cover  a  square,  multiplying  the  number  of  squares 
in  the  roof  by  3  will  give  the  number  of  bunches  of  shingles 
required. 

EXAMPLE.    A  roof  contains  12  squares. 

12  x  3  =  36  bunches,  or  9000  shingles. 

The  above  methods  give  a  few  more  shingles  than  are  actually 
needed.  Usually  some  are  broken  or  are  poor,  or  the  bunches  run 
a  little  short ;  hence  it  is  well  to  buy  a  few  extra,  to  be  sure  of  having 
enough. 


34  SHOP  PROBLEMS  IN  MATHEMATICS 

(4)  Where  roofs  are  straight  it  is  sometimes  convenient 
to  get  the  number  of  courses  and  multiply  by  the  number 
of  shingles  in  a  course. 

EXAMPLE.  The  length  of  the  ridge  of  a  roof  is  25  ft.  and 
the  rafters  are  15  ft.  on  either  side.  Required  the  number 
of  shingles. 

25  x  3  =  75  shingles  in  a  course. 

To  find  the  number  of  courses  (5  in.  to  the  weather), 
since  there  are  12  courses  for  every  5  ft.,  and  it  is  ,30  ft. 
over  the  roof,  or  6  times  5,  we  therefore  multiply  12  by  6, 
obtaining  72,  or  the  number  of  courses.  Hence 

25  x  3  x  (30  -5-  5)  x  12  =  5400  shingles. 

It  will  be  noticed  that  method  (4)  gives  the  exact  number  required. 
To  this  must  be  added  the  extra  required  for  waste,  etc. 

PROBLEMS 

1.  A  lean-to  roof  is  18  ft.  wide  and  15  ft.  long.    How 
many  shingles  laid  5  in.  to  the  weather  are  required  ?    (By 
method  (4).) 

2.  Find  the  number  of  shingles  required  for  a  roof,  if 
the  length  of  the  ridge  is  32  ft.,  the  length  of  the  rafters 
15  ft.,  and  the  shingles  are  laid  5  in.  to  the  weather.    (By 
method  (3).) 

3.  The  roof  of  a  bungalow  has  a  10-ft.  run,  a  5-ft.  rise, 
and  an  eaves  projection  of  2  ft.    If  the  ridge  of  the  house 
is  25  ft.  long,  find  how  many  shingles  are  required,  if  they 
are  laid  4  in.  to  the  weather.    (Use  method  (3).) 

4.  How  many  shingles  are  required  to  cover  a  hip  roof, 
if  the  building  is  25  ft.  square  and  the  rise  of  the  roof  is 
10  ft.,  the  shingles  to  be  laid  5  in.  to  the  weather  and  the 
eaves  projection  to  be  18  in.?    (Use  method  (2).) 

5.  A  cottage  has  four  gable  ends,  each  18  ft.  at  the 
base  and  9  ft.  high.    If  we  lay  the  shingles  4^  in.  to  the 


HOUSE  BUILDING  35 

weather,   how   many   are    required    to   cover  the   4    gable 
ends  ? 

6.  A  house  is  built  in  the  form  of  the  letter  T;  the  length 
of  the  ridge  of  the  top  of  the  T  is  35  ft.,  and  the  rafters  are 
17  ft.  long,  with  an  additional  eaves  projection  of  20  in. ; 
the  length  of  the  ridge  of  the  stem  of  the  T  is  20  ft.,  and 
its  rafters  also  are  17  ft.  long,  with  an  additional  20-in. 
eaves  projection.    If  the  width  of  each  part  of  the  house  is 
24  ft.,  how  many  shingles,  laid  5  in.  to  the  weather,  will 
cover  the  roof  ? 

7.  A  tower  10  ft.  square  has  a  hip  roof  with  a  10-ft. 
rise.    With  shingles  at  $4.50  per  M,  what  will  be  the  cost 
of  shingling  the  roof,  the  shingles  to  be  laid  5  in.  to  the 
weather,  the  roof  to  have  an  eaves  projection  of  24  in.,  and 
the  cost  of  nails  and  the  labor  of  laying  the  shingles  to  be 
$2.50  per  M  ?   (By  method  (3).) 

8.  An  octagonal  tower  is  12  ft.  across  between  any  two 
parallel  sides,  and  has  an  eaves  projection  of  20  in.    If  the 
rise  is  9  ft.  and  we  lay  the  shingles  5  in.  to  the  weather, 
how  many  are  required  for  the  roof  ?    (By  method  (2).) 

9.  A  square  house  32  ft.  on  a  side  has  a  veranda  with  a 
circular  corner  all  the  way  across  the  front  and  halfway 
down  the  side.    If  the  length  of  the  rafters  is  11  ft.  and 
we  lay  the  shingles  4  in.  to  the  weather,  how  many  are 
required  for  the  roof?    (By  method  (3).) 

10.  How  many  shingles  laid  5  in.  to  the  weather  will  be 
required  to  cover  the  roof  of  a  hexagonal  church  steeple 
which  is  6  ft.  on  a  side  at  the  base  and  30  ft.  high  ?    (By 
method  (3).) 

11.  A  round  tower  is  11  ft.  6  in.  in  diameter  and  the  rise 
of  the  roof  is  10  ft.    If  we  lay  the  shingles  5  in.  to  the 
weather,   how  much   will   shingles    for   the    roof   cost  at 
$4.50  per  M? 


36  SHOP  PROBLEMS  IN  MATHEMATICS 

12.  At  $5  per  M,  what  will  be  the  cost  of  special  4-in. 
shingles  for  the  gable  end  of  a  house  that  is  26  ft.  wide 
and  that  has  a  half-pitch  roof,  the  shingles  to  be  laid  5  in. 
to  the  weather  ? 

13.  A  pavilion  is  hexagonal  in  shape;  the  rise  of  the  roof 
is  10  ft.  and  the  length  of  the  hip  or  corner  rafter  is  18  ft. 
If  the  shingles  cost  $4.50  per  M  and  are  laid  5  in.  to  the 
weather,  what  will  be  the  cost  of  shingling  the  pavilion  if 
the  nails  and  labor  of  laying  costs  $2.50  per  M  ? 

BRICKWORK 

35.  Size  and  shape  of  bricks.   Bricks  vary  in  size  and  shape 
according  to  the  locality  in  which  they  are  made  and  the 
use  to  which  they  are  to  be  put. 

(1)  Ordinary  red  bricks  are  approximately  2^  in.  thick, 
3|  in.  wide,  and  8  in.  long.    In  estimating  the  number  of 
these  bricks  for  a  wall,  1\  bricks  are  generally  figured  per 
square  foot  of  wall  surface  1  brick  thick. 

(2)  The  so-called  "  Norman  "  bricks  are  2J  in.  thick  by 
4  in.  wide  by  12  in.  long,  and  are  figured  4J  bricks  per 
square  foot  of  wall. 

(3)  "  Roman  "  bricks  are  1£  in.  by  4  in.  by  12  in.,  and 
are  estimated  at  7  bricks  per  square  foot  of  wall. 

(4)  The  standard  face  bricks  are  2£  in.  thick,  4  in.  wide, 
and  8£  in.  long.    They  also  are  estimated  at  7  bricks  per 
square  foot  of  wall. 

36.  Measuring  brickwork.    Brickwork  is  generally  meas- 
ured by  the  square  foot,  square  yard,  etc.,  of  wall  (surface) 
so  many  bricks  thick.    For  example,  a  4-in.  wall  would  be 
1  brick  thick,  an  8-in.  wall  2  bricks  thick,  and  so  on.    Brick- 
work is  also  estimated  by  cubic  measure,  in  which  case  a 
wall  12  in.  thick  is  figured  to  contain  22  bricks  for  every 
square  foot  of  wall  face,  and  an  8-in.  wall  15  bricks.    This 


HOUSE  BUILDING  37 

can  be  done  only  when  the  bricks  throughout  the  wall  are 
of  uniform  size. 

When  openings  for  doors,  windows,  etc. ,  are  few  and  small  com- 
pared with  the  size  of  the  wall,  they  are  not  taken  into  account  in 
estimating  the  number  of  bricks.  If  large,  they  should  be  figured  out. 

In  ordinary  brickwork  the  amount  of  mortar  used  is  from  12  to 
18  bushels  per  1000  bricks. 

PROBLEMS 

1.  How  many  common  bricks  are  required  for  a  wall  8  ft. 
high,  26  ft.  6  in.  long,  and  2  bricks  thick  ? 

2.  A  12-in.  wall  is  faced  with  Norman  bricks  and  filled  in 
with  common  red  brick.  If  the  wall  is  18  ft.  7  in.  high  and  41 
ft.  6  in.  long,  how  many  of  each  kind  of  brick  are  required? 

3.  How  many  Roman  bricks  are  required  to  veneer  a 
a  house  1  brick  thick  up  to  the  second  floor,  the  size  of 
the  house  being  26  ft.  6  in.  by  35  ft.,  if  we  deduct  200 
sq.  ft.  for  door  and  window  openings  and  the  veneer  work 
is  10  ft.  high  ? 

4.  A  16-in.  wall  is  faced  on  both  sides  with  standard-face 
brick  and  filled  with  common  brick.    If  the  wall  is  60  ft. 
long  and  31£  ft.  high,  and  has  8  openings  for  windows  3|  ft. 
by  7  ft.,  how  many  of  each  kind  of  brick  are  required  ? 

5.  If  common  red  bricks  cost  $8  per  M,  how  much  will 
it  cost  to  lay  a  wall  18  ft.  high  by  49  ft.  long  by  12  in. 
thick,  the  labor  and  mortar  to  cost  $12  per  M? 

37.  Chimneys.  Common  brick  chimneys  are  generally  des- 
ignated as  so  many  bricks  long  and  so  many  bricks  wide, 
outside  measure.  It  usually  takes  from  4|  to  5  bricks  for 
1  ft.  in  height.  Hence,  to  find  the  number  of  bricks  required 
for  a  chimney,  multiply  the  number  of  bricks  in  one  course  by 
4t  or  5,  as  the  case  may  be,  and  this  number  again  by  the 
Iiei'jJit  of  the  chimneij  in  feet. 


38  SHOP  PROBLEMS  IN  MATHEMATICS 

PROBLEMS 

1.  How  many  bricks  are  required  for  a  chimney  2  bricks 
square  and  10  ft.  high  ?    (5  bricks  to  the  foot.) 

2.  A  chimney  is  2£  bricks  long  and  2  bricks  wide.    If  it 
is  35  ft.  high,  how  many  bricks  does  it  require  ?    (5  bricks 
to  the  foot.) 

3.  A  chimney  36  ft.  from  the  bottom  of  the  cellar  to  the 
roof  is  laid  of  common  red  bricks.    Above  the  roof  it  pro- 
jects 4£  ft.  and  is  laid  of  Roman  bricks.    If  it  is  5  com- 
mon bricks  long  and  2  common  bricks  wide  (40  in.  by  16  in.) 
and  has  3  flues  separated  by  2  brick  partitions  with  bricks 
placed  flatwise,  how  many  of  each  kind  of  brick  are  re- 
quired ?    (The  common  bricks  lay  5,  and  the  Koman  bricks 
7,  to  the  foot.) 

LATHING  AND  PLASTERING 

38.  Wood  laths.    Common  wood   laths   are  about   f  in. 
thick,  from  li  in.  to  1^  in.  wide,  and  4  ft.  long.    They  are 
generally  put  up  in  bunches  of  50  laths  each,  although  in 
some  localities  100  make  a  bunch.    Every  bunch  of  50  is 
figured  to  cover  about  25  sq.  ft.  Every  lath  requires  at  least 
4  nails,  and  there  are  about  480  3d  lath  nails  to  the  pound. 

39.  Plastering.    Plastering  is  usually  done  at  so  much 
per  square  yard  of  wall  surface. 

40.  Common  mortar.    Common  mortar  is  generally  com- 
posed of  five  parts  sand,  one  part  lime,  and  one  half  part 
hair.    A  bushel  of  common  mortar  will  cover  about  3  sq.  yd. 
with  two  coats. 

41.  Patent  mortar.    In  addition  to  common  mortar  there 
are  also  several  kinds  of  patent  mortar  now  in  use.    Their 
covering  capacities  vary  with  the  nature  of  their  materials 
and  the  kinds  of  work  on  which  they  are  used. 


HOUSE  BUILDING  39 

PROBLEMS 

1.  How  many  laths  will  it  take  to  cover  both  sides  of  a 
partition  that  is  9  ft.  high  and  14 J  ft.  long  ? 

2.  A  room  is  12  ft.  high,  20  ft.  6  in.  wide,  and  34  ft.  8  in. 
long.    How  many  laths  and  how  many  pounds  of  lath  nails 
does  it  require  for  the  overhead  and  side  walls,  if  we  deduct 
3  squares  for  openings  ? 

3.  The  first  floor  of  a  house  is  divided  into  4  rooms  as 
follows  :  12  by  14  ft.,  10  by  10  ft.,  8  by  10  ft.,  and  12  by 
12  ft.  respectively.    If  the  rooms  are  9  ft.  high  and  we 
take  out  350  sq.  ft.  for  door  and  window  openings,  how 
much  will  it  cost  to  lath  the  rooms,  the  laths  costing  $5 
per  1000,  the  nails  5/  per  pound,  and  the  work  of  putting 
on  25  cents  per  100  laths  ? 

4.  If  it  takes  about  18  laths  per  square  yard,  how  many 
laths  and  lath  nails  are  required  for  the  side  walls  of  a 
room  32  ft.  wide,  46  ft.  8  in.  long,  and  16  ft.  high,  there 
being  8  windows  and  4  doors,  averaging  25  sq.  ft.  each,  and 
a  wainscoting  all  around,  4  ft.  up  from  the  floor  ? 

5.  How  much  common  mortar  is  required  to  plaster  two 
coats  on  the  four  side  walls  and  ceiling  of  a  room  that  is 
12  ft.  wide,  13  ft.  long,  and  9  ft.  high,  if  we  deduct  100 
sq.  ft.  for  window  and  door  openings  ? 

6.  With  sand  at  $2  per  cubic  yard,  lime  at  $1.50  per 
barrel  containing  3  bushels,  and  hair  at  60  cents  per  bushel, 
what  will  be  the  cost  of  mortar  for  two-coat  work  for  the 
side  walls  and  ceilings  of  the  following  rooms  ? 

1  room  12£  ft.  by  16£  ft.  by  10  ft.  high. 
1  room  Hi  ft.  by  12  ft.  8  in.  by  10  ft.  high. 
1  room  9£  ft.  by  11  ft.  6  in.  by  10  ft.  high. 
Each  room  has  one  door  opening  3  ft.  by  7^  ft.  and 
two  windows  3  ft.  by  7  ft.  each. 


40  SHOP  PROBLEMS  IN  MATHEMATICS 

7.  The  price  of  a  certain  patent  plaster  is  $11  per  ton; 
it  has  a  covering  capacity  of  120  sq.  yd.  per  ton.     If  the 
mason  charges  15  cents  per  square  yard  for  labor  and  we 
deduct  400  sq.  ft.  for  door  and  window  openings,  what  will 
be  the  cost  of  plastering  the  side  walls  of  a  three-room 
summer  cottage,  the  sizes  of  which  are  as  follows  ? 

1  room  13£  ft.  by  23  ft.  by  11  ft.  high. 

1  room  10  ft.  by  12  ft.  by  11  ft.  high. 

.1  room  9  ft.  4  in.  by  12  ft.  by  11  ft.  high. 

2  closets  3  ft.  by  5  ft.  6  in.  by  11  ft.  high. 

8.  A  mason  takes  the  contract  to  furnish  all  materials 
and  do  the  work  involved  in  lathing  and  plastering  a  six- 
room  flat,  for  50  cents  per  square  yard.    The  sizes  of  the 
rooms  of  the  flat  are  as  follows  : 

1  room  81  ft.  by  12  ft.  by  9  ft.  high. 

1  room  15  ft.  by  11  ft.  4  in.  by  9  ft.  high. 

1  room  10  ft.  by  11  ft.  4  in.  by  9  ft.  high. 

1  room  13  ft.  by  9^  ft.  by  9  ft.  high. 

1  room  12  ft.  8  in.  by  15  ft.  by  9  ft.  high. 

1  room  7  ft.  by  11  ft.  by  9  ft.  high. 

1  hall  3  ft.  by  16  ft.  by  9  ft.  high. 

3  closets,  including  bathroom,  averaging  4  ft.  by  6  ft. 
by  9  ft.  high. 

If  100  sq.  yd.  are  deducted  for  door  and  window  openings, 
how  much  does  the  mason  receive  for  the  job  ? 

9.  If,  in  Problem  8,  the  laths  cost  $4  per  thousand,  the 
sand  $2  per  cubic  yard,  the  hair  60  cents  per  bushel,  the 
lime  $1.60  per  barrel,  containing  3  bushels,  the  lath  nails 
5  cents  per  pound,  with  450  nails  to  a  pound  and  4  nails 
to  every  lath,  the  labor  of  lathing  25  cents  per  100,  and 
the  labor  of  plastering  for  the  two  coats  18  cents  per 
square  yard,  does  the  mason  make  or  lose  money,  and 
how  much  ? 


HOUSE  BUILDING  41 

STONEWORK,  EXCAVATING,  CAPACITIES  OF  BINS 

42.  Approximate  cubical  contents.  The  following  method 
is  useful  in  finding  the  cubical  contents  of  rectangular 
solids  where  a  few  inches  may  be  disregarded,  as  in  the 
excavating  of  earth,  in  brickwork,  cementwork,  certain 
kinds  of  stonework,  etc. 

For  finding  the  area  of  one  surface  of  the  solid  see  sect.  19. 
Multiply  this  area  by  the  other  dimension  of  the  solid  in 
the  same  manner  as  before. 

EXAMPLE.  A  rectangular  solid  is  3  ft.  7  in.  wide,  4  ft. 
6  in.  long,  and  2  ft.  8  in.  thick.  What  is  its  cubical  con- 
tents ? 

ACTUAL  WORK  EXPLANATION 

3  ft.        7  in.  4  ft.  x  3  ft.  =  12  sq.  ft. 

4 6_  4  ft.  x  7  in.  =  f  |  sq.  ft. 

12  =2  sq.  ft.  and  T4^  sq.  ft. 

2  4  6  in.  x  4  ft.  =  j|  sq.ft. 

16  =  1  sq.  ft.  and  ,T\  sq.  ft. 

3     6  6in.x7in.  =  42sq.  in. 

16  2  =  T35  sq.ft.  and  T|?  sq.ft. 

28  6  in  the  third  column  is  called  1 

32  and  added  to  the  second  column, 

4  giving  a  sum  of  j|  sq.  ft.  —  1  sq. 

10  8  ft.  and  T\  sq.  ft.  Adding  the  1  sq. 

1  ft.  to  the  first  column,  we  have 

43  cu.  ft.  TV  cu.  ft.         16  sq.  ft. 

2  ft.  x  16  sq.  ft.  =  32  cu.  ft. 
2  ft.  x  T22  sq.  ft.  =  T\  cu.  ft. 
8  in.  x  16  sq.  ft.  =  T\  ft.  x  16  sq.  ft.  =  iff.  cu.  ft. 

=  10  cu.  ft.  and  T8^  cu.  ft. 
8  in.  x  T\  sq.  ft.  =  T\  ft.  x  T\  sq.  ft.  =  T^  cu.  ft. 

=  T^  cu.  ft.  and  T|3  cu.  ft. 


42  SHOP  PROBLEMS  IN  MATHEMATICS 

Adding  the  second  column,  we  have  jf  cu.  ft.  =  1  cu.  ft. 
and  yL  cu.  ft. 

Adding  1  cu.  ft.  to  the  first  column,  the  result  is  43 TT^ 
cu.  ft. 

PROBLEMS 

By  the  above  method  find  the  approximate  cubical  contents 
of  rectangular  solids  having  the  following  dimensions : 

1.  1  ft.  6  in.  by  1  ft:  4  in.  by  2  ft. 

2.  1  ft.  4  in.  by  2  ft.  3  in.  by  2  ft.  2  in. 

3.  2  ft.  5  in.  by  2  ft.  3  in.  by  3  ft.  3  in. 

4.  4  ft.  7  in.  by  5  ft.  9  in.  by  3  ft.  3  in. 

5.  6  ft.  7  in.  by  7  ft.  9  in.  by  4  ft.  10  in. 

6.  3  ft.  3  in.  by  6  ft.  5  in.  by  12  ft.  11  in. 

7.  20  ft.  6  in.  by  30  ft.  8  in.  by  9  ft.  9  in. 

8.  30  ft.  4  in.  by  19  ft.  11  in.  by  10  ft.  2  in. 

9.  36  ft.  7  in.  by  27  ft.  8  in.  by  12  ft.  9  in. 

10.  23  ft.  5  in.  by  18  ft.  7  in.  by  8  ft.  7  in. 

11.  How  many  cubic  feet  of  stone  mason  work  in  a  wall 
16  in.  thick,  9  ft.  3  in.  high,  and  40  ft.  6  in.  long  ? 

12.  If  a  ton  of  coal  contains  35  cu.  ft.,  how  many  tons  will  a 
bin  4  ft.  3  in.  wide,  20  ft.  4  in.  long,  and  5  ft.  8  in.  high  hold  ? 

13.  At  $6.75  per  ton,  how  much  will  a  car  of  anthracite 
coal  cost  if  the  car  is  3  ft.  3  in.  deep,  8  ft.  1  in.  wide,  and 
33  ft.  long  ? 

14.  A  man  wishes  to  build  a  potato  bin  that  will  hold 
40  bushels.    It  is  to  be  6  ft.  8  in.  long  and  3  ft.  6  in.  high, 
and  a  bushel  contains  1|  cu.  ft.    What  must  be  the  width 
of  the  bin  ? 

15.  Susquehanna  granite  weighs  approximately  169  Ib. 
per  cubic  foot.    What  will  be  the  weight  of  a  block  that  is 
4  ft.  6  in.  long,  3  ft.  8  in.  wide,  and  2  ft.  5  in.  thick  ? 


NERAL  CONSTRUCTION  43 


16.  How  many  cubic  yards  in  a  bank  of  earth  12  ft.  8  in. 
wide,  21  ft.  6  in.  long,  and  7  ft.  6  in.  deep  ? 

17.  A  cellar  wall  is  23  ft.  6  in.  wide  by  35  ft.  8  in.  long, 
outside  measure,  and  8  ft.  6  in.  high.    If  the  wall  is  16  in. 
thick,  how  many  cubic  yards  of  mason  work  does  it  contain  ? 

18.  If  a  bushel  of  potatoes  weighs  60  Ib.  and  contains 
approximately  li  cu.  ft.,  how  many  bushels  can  be  loaded 
into  a  car  that  is  33  ft.  long  and  8  ft.  2  in.  wide,  inside 
measure,  the  potatoes  to  be  piled  6  ft.  6  in.  high  ?    What 
will  be  the  weight  of  the  carload  ? 

19.  The  depths  of  earth  taken  in  various  points   of  a 
cellar  that  is  to  be  excavated  are  4  ft.  2  in.,  5  ft.  6  in.,  7  ft., 
6  ft.  6  in.,  3  ft.  10  in.,  7  ft.  4  in.,  8  ft.  8  in.,  and  9  ft.  re- 
spectively.  If  the  cellar  is  to  be  60  ft.  6  in.  wide  and  90  ft. 
10  in.  long,  how  many  cubic  yards  of  earth  must  be  re- 
moved ?    (Take  the  average  depth  of  the  cellar.) 

20.  At  $1  per  cubic  yard,  what  will  be  the  cost  of  blasting 
out  the  rock  from  a  cellar  that  is  22  ft.  6  in.  wide  by  46  ft. 
6  in.  long,  the  depth  of  the  rock  at  various  points  being  4  ft., 
2  ft.  6  in.,  3  ft.  4  in.,  6  ft.  8  in.,  7  ft.,  5  ft.,  and  4  ft.  6  in.  ? 

43.  Amount  of  lumber  for  bins,  boxes,  and  furniture.  Before 
working  out  the  following  problems  the  student  should 
read  Chapter  XVI. 

PROBLEMS 

1.  A  man  wishes  to  build  a  coal  bin  5  ft.  deep  and  7  ft. 
square  on  the  outside,  each  side  to  be  stayed  by  three  2-  by 
4-in.  pieces,  6  ft.  long,  and  the  bottom  to  be  laid  on  four 
pieces  of  2-  by  4-in.  stock,  7  ft.  long.   If  the  boards  used  are 
1  in.  thick,  how  many  feet  of  lumber  are  required  for  the  bin? 

2.  If  there  are  35  cu.  ft.  of  anthracite  coal  in  a  ton,  how 
much  1-in.  lumber  will  it  take  to  line  a  coal  bin  to  hold 
6  tons,  the  width  of  the  bin  to  be  7  ft.  and  its  length  8  ft.  ? 


44  SHOP  PROBLEMS  IN  MATHEMATICS 

3.  A  man  wishes  to  put  up  a  billboard  8  ft.  wide  and 
30  ft.  long,  using  matched  lumber  £  in.  thick.    The  frame- 
work for  the  board  consists  of  2  pieces  of  2-  by  4-in.  stock, 
running  lengthwise,  and  6  pieces  of  2-  by  4-in.  stock,  run- 
ning crosswise  or  up  and  down.    If  he  makes  no  allowance 
for  waste  on  the  scantling  and  adds  25  ft.  for  waste  and 
squaring  to  the  matched  stuff,  how  much  of  each  kind  of 
lumber  must  he  buy  for  the  board  ? 

4.  A  grain  bin  with  a  hopper  bottom  is  lined  with  1-in. 
boards  of  double  thickness ;  the  bin  is  16  ft.  deep  down  to 
the  hopper  part  and  is  13  ft.  square ;  the  hopper  has  a  per- 
pendicular height  of  5  ft.    If  we  allow  25  ft.  for  waste  in 


FIG.  21.    HOPPER 

cutting  the  boards,  how  much  lumber  is  required  for  the 
bin  ?    (See  Fig.  21.) 

5.  A  manual-training  bench  top  is  15  in.  wide,  4  ft.  long, 
and  2|  in.  thick.    To  prevent  warping  it  is  constructed  of 
1^-in.  by  3-in.  pieces  (in  the  rough  stock)  glued  together. 
Allowing  £  in.  both  on  the  thickness  and  width  and  1  in. 
on  the  length  of  each  piece  for  truing  up,  how  much  lum- 
ber is  required  for  the  top  ?    (See  Fig.  22.) 

6.  A  cabinet  box  is  4  in.  high,  6  in.  wide,  and  10  in.  long, 
inside  measure.    If  the  side  and  end  pieces  are  constructed 
of  stock  |  in.  thick,  and  the  top  and  bottom  of  stock  f  in. 
thick,  how   much   stock   is   required  for  its   construction, 


GENERAL   CONSTRUCTION  45 

allowing  \  in.  on  the  width  and  f  in.  on  the  length  of 
every  piece  for  rough  stock  ? 

7.  How  much  lumber  is  necessary  to  construct  a  mission 
umbrella  rack  having  4  corner  posts  If  in.  square  and  2  ft. 
long,  2  crosspieces  on  each  of  the  4  sides  J  by  2  by  12  in., 
and  a  £-in.  board  in  the  bottom,  allowing  ^  in.  on  the  width 
and  thickness  and  1  in.  on  the  length  of  every  piece  for 
rough  stock  ? 

8.  A  quartered-oak  mission  table  top  is  1^  in.  thick,  30  in. 
wide,  and  47  in.  long  when  finished.    If  it  is  constructed  of 
four  boards  jointed  edge  to  edge,  how  many  feet  does  it 
contain,  allowing  at  least  £  in.  on  the  width  of  every  piece 


FIG.  22.    BENCH  TOP 

for  jointing  and  1  in.  on  the  length  of  every  piece  for  tru- 
ing up  ?    What  size  of  boards  would  you  select  ? 

9.  With  oak  lumber  at  $75  per  M,  what  will  be  the  cost 
of  lumber  for  36  footstools  12  by  16  by  8  in.  high,  the  rough 
stock  for  every  stool  to  be  as  follows  ? 

4  legs  2  by  2  by  9  in.  long. 
2  rails  J  by  3  by  15  in.  long. 
2  rails  J  by  3  by  11  in.  long. 
This  does  not  provide  for  the  top,  which  is  usually  upholstered. 

10.  A  threefold  screen  is  58  in.  high  and  54  in.  wide.  Every 
panel  has  2  wooden  stiles  and  2  rails :  the  stiles  are  1  in. 
by  If  in. ;  the  bottom  rails  are  1  in.  by  4  in.;  and  the  top 
rails  are  1  in.  by  If  in.  when  finished.    If  the  rails  go  into 


46 


SHOP  PROBLEMS  IN  MATHEMATICS 


the  stiles  with  a  tenon  1  in.  long,  state  the  number  of  pieces 
required,  the  size  of  every  piece  (rough  stock),  and  the  num- 
ber of  feet  of  lumber  required  for  the  screen. 

11.  With  ash  lumber  at  $85  per  M,  how  much  will  the 
wood  cost  for  a  porch  swing  24  in.  high,  24  in.  wide,  and 
60  in.  long,  the  rough  stock  to  be  as  follows  ? 

4  posts  3  by  3  by  25  in.  long. 
2  rails  1£  by  4J  by  26  in. 

2  rails  1^  by  4^  by  62  in. 
2  rails  Ij-  by  3£  by  26  in. 
10  bottom  pieces  -J  by  6  by  22  in. 

5  slats  for  back  and  side  -j-  by  10  by  15  in. 

12.  A  quartered-oak  library  table  is  45  in.  long,  30  in. 
wide,  and  30  in.  high  ;  the  top  when  finished  is  1J  in.  thick  ; 

the  legs  are  2J  in. 
square  ;  the  4  rails 
(2  side  and  2  end) 
are  J  by  5f  in. ;  the 
legs  set  back  from 
the  end  of  the  top 
3  in.  and  from  the 
side  If  in. ;  the  rails 
go  into  the  legs 
with  a  IJ-in.  tenon. 
Required  the  size  of 
every  piece  of  rough 
stock,  the  number 
of  pieces,  and  the 
number  of  feet  of  lumber  (the  top  to  contain  4  pieces). 

13.  At  $100  per  M,  what  will  quartered-oak  lumber  cost 
for  10  armchairs,  to  be  constructed  as  follows  : 

Size  of  chairs,  31  in.  high,  25  in.  deep,  and  26  in.  wide. 
To  have  4  posts  2J  in.  square  when  finished  full  height. 


FIG.  23.    ARMCHAIR 


GENERAL   CONSTRUCTION  47 

To  have  4  rails  1  in.  by  5£  in.  when  finished. 

To  have  3  rails  1  in.  by  2£  in.  when  finished,  one  at 
the  top  of  the  back  and  one  at  the  top  of  either  side. 

To  have  4  slats  for  side  and  back  £  in.  by  6  in.  by  15  in. 
long  (rough  stock).  (See  Fig.  23.) 

14.  What  is  the  smallest  rectangular  piece  of  stock  from 
which  a  hexagonal  jardiniere-stand  top  6  in.  on  a  side  can 
be  cut,  allowing  ^  in.  on  the  width  and  1  in.  on  the  length 
of  the  piece  for  rough  stock  ? 

15.  A  corner  shelf  is  16  in.  on  each  of  its  shorter  sides. 
What  is  the  smallest  piece  of  rectangular  stock  from  which 
it  can  be  cut,  allowing  \  in.  on  the  width  and  1  in.  on  the 
length  for  rough  stock  ? 

16.  A  boy  wishes  to  construct  an  octagonal  table  top 
12  in.  on  a  side.    What  size  of  board  must  he  buy,  the  top 
to  contain  4  pieces  and  to  be  1  in.  thick  when  finished  ? 
Allow  \  in.  on  the  width  of  every  piece  and  1  in.  on  the 
length  for  jointing  and  truing  up. 

17.  How  much  lumber  is  required  for  a  magazine  cabinet 
40  in.  high,  14  in.  wide,  and  10  in.  deep  when  finished,  the 
rough  stock  to  be  as  follows  : 

2  uprights  41  in.  high  by  10^  in.  wide  by  1  in.  thick. 
4  shelves  15  in.  long  by  9£  in.  wide  by  1  in.  thick. 
2  cross  rails  15  in.  long  by  4  in.  wide  by  1  in.  thick. 

44.  Finding  a  square  equivalent  to  two  squares.  To  find  the 
side  of  a  square  that  has  an  area  equal  to  the  areas  of  two 
given  squares,  denote  the  sides  of  the  given  squares  by  s 
and  «!  and  the  side  of  the  required  square  by  S,  as  in 
Fig.  24. 

If  a  right  triangle  is  constructed  with  the  legs  equal 
respectively  to  s  and  slt  the  hypotenuse  S  will  be  the  side 
of  the  required  square,  since  S2  =  s2  +  sf. 


48 


SHOP  PROBLEMS  IN  MATHEMATICS 


This  calculation  may  be  made  by  any  of  the  following 
methods : 

(1)  On  the  steel  square,  laying  off  s  and  sx  on  the  sides 
of  the  square  and  measuring  the  diagonal  S  from  the  ends 

of  s  and  slf  (See  Fig.  16.) 

(2)  On    graph    or    cross- 
section    paper,    using     any 
convenient  unit. 

(3)  Drawing    to    scale   a 
right  triangle,  given  the  two 

legs  s  and  sx  and 
measuring  S. 

(4)  By  the  form- 
ula S2  =  s2  +  s?, 
substituting  the 
given  values  of  s 
and  sx  and  solv- 
ing for  S. 

EXAMPLE.  Two 
grain  bins  of  the 
same  depth  have 
bottoms  9  ft.  and 

FIG.  24.   SQUARE  EQUIVALENT  TO  Two  SQUARES 

spectively.  What 

must  be  the  size  of  the  square  bottom  of  a  third  bin  of  the 
same  depth  that  will  hold  as  much  as  the  two  given  bins  ? 


Sl  =  11. 

Substituting  in  the  formula  S2  —  s2  -f-  sf, 
S2  =  81  +  121  =  202. 
S  =  14.21  ft.  =  14  ft.  3  in. 

45.  Finding  a  circle  equivalent  to  two  circles.    Since  the 
areas  of  circles  are  to  each  other  as  the  squares  of  their 


GENERAL  CONSTRUCTION 


49 


diameters,  the  methods  of  section  44  may  be  used  to  find 
the  diameter  of  a  circle  that  is  equivalent  to  two  given 
circles. 

Denoting  the  given  diameters  by  d  and  dly  and  the  re- 
quired diameter  by  Z>,  we  have  the  formula  D2  =  d2  -f-  d?, 
from  which  D  may  be  found  by  any  of  the  methods  of 
section  44. 

EXAMPLE.  Two  branches  of  an  iron  pipe  are  respectively 
2  in.  and  3  in.  in  diameter.  What  must  be  the  diameter  of 
the  pipe  into  which  they  empty,  in  order  that  the  water  may 
be  carried  off  if  the  velocity 
of  the  water  in  all  three 
pipes  is  the  same  ? 


FIG.  25.    CIRCLE  EQUIVALENT  TO 
Two  CIRCLES 


Substituting  in  the  form- 
ula D2  =  d2  +  d}, 

D2  =  4  +  9  =  13. 
D  =  3.6  in. 

Since  the  nearest  standard 
size  of  iron  pipe  above  3.6  in. 
is  4  in.,  the  result  should  be 
called  4  in. 

46.  Finding  a  square  equivalent  to  the  difference  of  two 
given  squares,  or  a  circle  equivalent  to  the  difference  of  two 
given  circles.  The  methods  of  sections  44  and  45  may  be 
used  when  s,  s1?  d,  or  d1  is  the  unknown  quantity.  If  the 
work  is  done  by  drawing  to  scale,  as  in  Fig.  25,  a  right 
triangle  must  be  constructed,  given  the  hypotenuse  and 
one  side. 

If  the  formula  is  used, 

C2  _    C-2  C2 

s  —  o    —  «ij 

or  d'2  =  D2  —  d*. 


50  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE.  A  lead  water  pipe  2  in.  in  diameter  discharges 
into  a  tank.  The  water  leaves  the  tank  by  two  lead  pipes, 
one  of  which  is  \  in.  in  diameter.  What  must  be  the  diame- 
ter of  the  other  in  order  that  the  water  may  leave  the  tank 
as  fast  as  it  runs  in,  if  the  velocity  of  the  water  in  all  three 
pipes  is  the  same? 


d'2  =  D2  -  d}  =  4  -  1  =  3. 
d  =  1.7  in. 

Since  the  nearest  standard  size  of  lead  coil  pipe  above 
1.7  in.  is  If  in.,  the  result  should  be  called  If  in. 

SOME  STANDARD  SIZES  OF  PIPE 

Lead  coil  pipe      j,  ^,  >,  |,  |,  1,  li,  1|,  If,  2. 
Lead  waste  pipe  U,  2,  3,  3J,  4,  4-|,  5,  6. 

Iron  pipe  |,  1,  f  ,  |,  f  ,  1,  li,  1J,  2,  2J,  3,  3J,  4,  4J,  5,  6,  7, 

8,  9,  10,  11,  12. 

PROBLEMS 

1.  The  two  branches  of  an  iron  sewer  pipe  are  respec- 
tively 2£  and  3|  in.  in  diameter.    What  must  be  the  diame- 
ter of  the  iron  pipe  into  which  they  empty,  in  order  that 
the  sewage  may  be  carried  off  ? 

2.  How  long  must  the  side  of  a  square  plot  of  ground  be, 
to  contain  as  much  land  as  two  other  square  plots  whose 
sides  measure  100  ft.  and  150  ft.  respectively  ? 

3.  Find  the  diameter  of  a  circular  flower  bed  that  will 
be  equal  in  area  to  two  circular  flower  beds  having  diame- 
ters of  15  ft.  and  20  ft.  respectively. 

4.  A  lead  water  pipe  2  in.  in  diameter  discharges  into  a 
tank.    The  water  leaves  the  tank  by  two  lead  waste  pipes, 
the  diameter  of  one  of  which  is  1    in.    What  must  be  the 


GENERAL  CONSTRUCTION 


51 


diameter  of  the  other  pipe  in  order  that  the  water  may 
leave  the  tank  as  fast  as  it  runs  in  ? 

5.  A  vessel  12  in.  deep  and  15  in.  in  diameter  holds  about 
one  bushel.    Each  of  two  other  vessels  is  12  in.  deep,  the 
diameter  of  one  being  7  in.    What  must  be  the  diameter  of 
the  other  in  order  that  the  two  together  may  hold  as  much 
as  the  first  ? 

6.  Two  grain  bins  are  8  ft.  and  7  ft.  square  on  the  bottom 
respectively.    What  must  be  the  size  of  the  square  bottom 
of  the  third  in  order  that  it  may  hold  as  much  as  the  other 
two,  the  depth  of  all  three  being  the  same  ? 

7.  Water  is  conducted  into  a  tank  through  two  lead  coil 
pipes  having  diameters  of  f  in.  and  If  in.  respectively.   Find 
the  standard  size  of  the  lead  waste  pipe  that  will  allow  the 
water  to  run  out  as  fast  as  it  runs  in. 

If  all  pipes  are  iron,  supply  the  missing  -diameters  indi- 
cated in  the  following  table  : 

D  d  (fi 


8. 
9. 
10. 
11. 

i 

H 

2* 

41 

** 

2£ 

10 

3| 

12.  Two  branch  iron  sewer  pipes,  each  6  in.  in  diameter, 
empty  into  a  third  pipe ;    farther  down  the  line  another 
branch  8  in.  in  diameter  runs  into  the  sewer.    What  must 
be  the  diameter  of  the  iron  pipe  below  this  point  in  order 
to  carry  off  the  sewage? 

13.  The  side  of  a  square  plot  of  ground  measures  10  rd. 
Find  the  side  of  a  square  plot  twice  as  large.   Show  that  the 
ratio  of  the  side  of  the  second  plot  to  the  side  of  the  first 
plot  is  V2  =  1.414. 


52 


SHOP  PROBLEMS  IN  MATHEMATICS 


HEIGHTS  OF  TREES  OR  OTHER  OBJECTS 

47.  First  method.    By  the  isosceles  right  triangle.    The  ob- 
server stands  with  his  eye  at  A.    ABC  is  an  isosceles  right 
triangle.    Evidently  h,  the  height  of  the  tre,e,  is  equal  to  d, 
the  distance  from  the  observer  to  the  tree,  plus  a,  the  height 
of  the  observer's  eye  above  the  ground  (Fig.  26). 

WRITTEN    EXERCISE 

1.  If  a  =  6,  d  =  50',  find  h. 

2.  If  a  =  5'  6",  d  =  45'  9",  find  h. 

3.  If  a,  =  5i,  d  =  90§,  find 

48.  Second  method.    If  a 
triangle  is  used  in  which 
BC  =  2  AC,    show 

that  h  =  2  d  +  a. 


FIG.  26.   HEIGHTS  or  TREES 
FIRST  METHOD 


WRITTEN    EXERCISE 

1.  If  a  =  5'  4",  d  =  30',  find  h. 

2.  If  a  =  5'  3",  d  =  25',  find  h. 

3.  If  a  =  5£f,     d  =  42'  6",  find  h. 


HEIGHTS  OF  TREES 


53 


49.  Third  method.   To  find  h,  the  height  of  the  tree,  meas- 
ure the  shadow  d,  set  up  a  stake  h'f  measure  its  shadow  d', 

and  then,  using  the  proportion  -  =  — ,  solve  for  h. 

ct       ct 

EXAMPLE.    Find  hiid  =  30',  h'  ==  5',  d'  =  6'. 


, 

r          h      5 

/ 

30~6' 

/ 

;          5   X    30 



/ 

6                        / 

=  25. 

/ 

4 

/ 

/ 

«c 

/ 

/ 

/ 

/ 

AL.  1C 

•1       /               1 

/          f         / 

/.  *         / 

, 

!                                             .__*._. 

\*d'4      1*-    -  ^  J 

FIG.  27.    SECOND  METHOD 


FIG.  28.    THIRD  METHOD 


WRITTEN   EXERCISE 

Find  h  if  d,  h',  and  d'  have  the  values  indicated. 
d  h'  d' 


1. 

45'  3" 

5' 

3'  4" 

2. 

42'  6" 

5' 

6'  8" 

3. 

22'  4" 

5' 

10'  3" 

4. 

21'  5" 

5' 

15'  6" 

54 


SHOP  PROBLEMS  IN  MATHEMATICS 


5.  The  shadow  of  a  stake  driven  into  the  ground  perpen- 
dicularly is  2  ft.  and  the  height  of  the  stake  is  3  ft.    If  the 
shadow  of  a  near-by  tree  is  70  ft.  long,  how  high  is  the 
tree? 

6.  The  shadow  of  a  tree  is  40  ft.  long.    If  the  shadow  of 
a  stake  4  ft.  high  is  2  ft.  3  in.,  what  is  the  height  of  the  tree  ? 

50.  Fourth  method.  Set  two 
poles  in  a  line  with  the  tree 
and  perpendicular  to  the 
earth,  as  in  Fig.  29.  With 
the  eye  at  E,  a  point  on  one 
pole,  sight  across  the  other 
pole  to  the  base  and  to  the 
top  of  the  tree.  Have  an  £ 
assistant  mark  the  points 

where  the  lines  of  vision  cross 

/•\ 

the  second  pole  at  P  and  P'.  FIG.  29.    FOURTH  METHOD 

Measure  EP,  EA,  and  PP'. 


Then  A,  the  height  of  the  tree  = 


EA  x 


EP 


WRITTEN   EXERCISE 

Find  h  if  EA,  PP',  and  EP  have  the  values  indicated. 
EA  PP'  EP 


1. 

44' 

3' 

2' 

2. 

33' 

5'  6" 

2'  3" 

3. 

52'  6" 

5'  3" 

5' 

4. 

24'  4" 

5'  8" 

2'  4" 

5. 

25'  8" 

6' 

1'6" 

6. 

40'  3" 

5'  4" 

3' 

7. 

38'  5" 

5'  10" 

4'  3" 

HEIGHTS  OF  TREES 


55 


51.  Fifth  method.    The  observer  walks   on  level 
ground  to  a  distance  TE  from  the  foot  of  the 
tree,  about  equal  to  its  estimated  height  TT'. 
He  then  lies  on  his   back,  stretched  at          S* 
full   length,  with  his   eye   at  E  and 
his  feet  at  A.    An  assistant  marks 
the  point  B  on  a  perpendicular          ,-'' 
staff  erected  at  his  feet,          ,-'' 
the  exact  point  where 
his  line  of  vision 
to  the  top  of  the     E  -'-'--- 
tree   crosses  the  FlG-  30-   FlFTU  METHOD 

staff.  The  distances  BA,  EA,  and  ET  are  measured. 

BA  X  ET 


Then 


TT'  = 


EA 


EXERCISE 

Find  TT'  in  the  following  problems  : 
BA  ET  EA 


I. 

4'  2" 

52'  4" 

6' 

2. 

5'  3" 

45'  2" 

5'  10" 

3. 

8'  4" 

30'  5" 

5'    6" 

4. 

10'  6" 

23'  7" 

5'    4" 

52.    Sixth  method.    The  followin; 
method  gives  a  rough  approximation 
to  the  height  of  a  tree  : 

The  distance  ylZ?  is  marked       /'  ,. 
on  the  tree  as  high  as  the 
woodman  can  reach.  / 

Then,  stepping  back,     E^l 
a  pencil  PP'  is  held 
vertically  with   the  FIG.  31.    SIXTH  METHOD 


56 


SHOP  PROBLEMS  IN  MATHEMATICS 


eye  at  E,  as  in  the  figure.  The  pencil  is  raised  with  .P' 
transferred  to  P,  and  the  point  C  marked  with  the  eye. 
This  is  repeated  till  the  top  of  the  tree  T  is  reached. 
d,  d',  d",  d"',  div  are  approximately  equal. 

The  same  method  applies  to  finding  the  approximate 
diameter  of  the  tree  at  any  given  point  above  the  ground. 
From  the  height  and  diameter  a  rough  approximation  can 
be  made  of  the  amount  of  lumber  in  the  tree. 

ORAL    EXERCISE 

1.  If  d  =  6'  6"  and  the  pencil  is  raised  8  times  before 
the  top  of  the  tree  is  reached,  find  the  height  of  the  tree. 

If  n  is  the  number  of  times  the  pencil  is  raised,  find  the 
height  of  the  trees  in  the  following  problems : 


2. 

6'  2" 

6 

3. 

6'  6" 

8 

4. 

r-l 
i 

7 

5. 

7'  3" 

4 

NOTE.  For  a  description  of  instruments  used  in  forestry  for  meas- 
uring heights  of  trees,  the  student  is  referred  to  "The  Woodman's 
Handbook,"  Part  I,  by  H.  S.  Graves,  published  as  Bulletin  No.  36  by 
the  U.  S.  Dept.  of  Agriculture,  Bureau  of  Forestry. 

53.  To  measure  the  distance  across  a  river  gorge,  etc.    For 

the  principle  of  right  triangles  the  student  is  referred  to 
sections  44-46. 

PROBLEMS 

1.  Two  trees,  T  and  C,  stand  directly  opposite  each  other 
on  the  banks  of  the  river.  A  man  takes  a  right-angled  tri- 
angle ACS,  with  equal  sides  AC  and  BC,  and  marks  the 
line  CD  on  the  bank  at  right  angles  to  CT.  Then  he  walks 


DISTANCE  ACROSS  A  RIVER 


5T 


along  the  line  CD  until,  placing  the  triangle  to  his  eye  in  a 
horizontal  position,  he  sees  in  range  of  the  line  of  the  hy- 
potenuse A'B1  the  tree  Ton  the  opposite  bank,  and  in  range 
of  the  line  of  the  side  A'C'  the  tree  C  on  the  same  side  as 
himself.  If  the  distance  from  where  he  stands  to  the  tree 
on  the  same  side  is  700  ft.,  how  far  is  it  across  the  river  ? 


FIG.  32.    DISTANCE  ACROSS  A  RIVER 

2.  A  man  wishes  to  know  the  distance  across  a  gorge. 
He  drives  a  stake  on  one  bank  and  takes  for  his  observa- 
tion point  on  the  other  bank  a  round  stone.    He  then  takes 
a  right-angled  triangle,  the  sides  of  which  are  as  2  is  to  9, 
places  it  in  a  position  similar  to  that  of  Problem  1,  and 
walks  up  the  bank  until  the  hypotenuse  ranges  with  the 
stone  on  the  opposite  bank  and  the  short  side  ranges  with 
the  stake.    If  the  distance  from  where  he  stands  to  the 
stake  is  46  ft.,  what  is  the  width  of  the  gorge  ? 

3.  With  a  right-angled  triangle  with  sides  as  3  is  to  7J, 
placed  as  in  Problem  1,  we  find  the  distance  from  our  start- 
ing point  on  the  bank  of  a  river  to  a  tree  on  the  opposite 
side  to  be  460  ft.   We  take  a  stand  at  the  starting  point  and 
with  two  straight,  thin  strips,  get  the  range  of  the  top  of 


58  SHOP  PROBLEMS  IN  MATHEMATICS 

the  tree  with  one  of  the  strips  and  the  range  of  the  bottom 
with  the  other,  thus  forming  an  acute  angle  at  the  eye  with 
the  strips.  We  now  measure  a  distance  on  the  lower  strip 
(say  3  ft.  from  the  angle)  and  the  perpendicular  from  this 
point  to  the  upper  strip,  which  we  find  to  be  3  in.  What 
is  the  height  of  the  tree  on  the  opposite  bank  ? 

HEIGHTS  OF  CLIFFS 

54.  By  falling  bodies.  The  number  of  feet  passed  over  by 
a  body  falling  from  a  state  of  rest  is  equal  to  16.08  times  the 
square  of  the  number  of  seconds  during  which  the  body  falls. 

(1)  Denoting  this  space  by  S  and  the  number  of  seconds 
by  t,  write  the  formula  for  S  in  terms  of  t. 

(2)  A  stone  dropped  from  a  cliff  reaches  the  ground  in 
4  seconds.    Find  the  height  of  the  cliff. 

(3)  From  a  point  on  the  Palisades  a  stone  is  dropped, 
reaching  the  river  in  3  seconds.    How  high  is  the  point 
above  the  river  ? 

(4)  A  stone  dropped  from  the  top  of  a  building  reaches 
the  ground  in  2f  seconds.    Find  the  height  of  the  building. 

MISCELLANEOUS   PROBLEMS 

1.  A  man  holds  a  ruler  vertically  between  his  thumb 
and  finger  at  arm's  length,  and  places  it  in  such  a  position 
that  he  can  see  the  top  of  a  tree  over  the  top  of  the  ruler, 
and  the  bottom  of  the  tree  on  a  line  with  the  top  of  his 
thumb.    He  finds  the  distance  from  his  eye  to  the  top  of 
his  thumb  to  be  2  ft.  4  in.,  and  the  distance  the  ruler  pro- 
jects above  his  thumb  to  be  9£  in.    If  from  the  base  of  the 
tree  to  where  he  stands  the  distance  is  60  ft.,  how  high  is 
the  tree  ? 

2.  What  would  be  the  height  of  a  church  steeple,  accord- 
ing to  the  above  method,  if  the  distance  from  the  top  of 


MISCELLANEOUS  PROBLEMS  59 

the  ruler  to  the  top  of  the  man's  thumb  is  2  ft.,  the  dis- 
tance from  his  eye  to  the  top  of  the  thumb  is  1  ft.  9  in.,  and 
the  distance  from  his  eye  to  the  ground  directly  below  the 
steeple  is  100  ft.  ? 

3.  The  height  of  a  building  on  one  bank  of  a  river  is 
55  ft.    By  holding  a  ruler  on  the  opposite  bank  according 
to  Problem  1,  we  find  when  the  top  of  the  ruler  ranges  with 
the  top  of  the  building,  and  the  top  of  the  thumb  with  the 
bottom  of  the  building,  the  ruler  projects  above  the  thumb 
6|-  in.,  and  the  distance  from  the  thumb  to  the  eye  is  23  in. 
What  is  the  width  of  the  river  ? 

4.  The  legs  of  a  right-angled  triangle  are  to  each  other 
as  3  is  to  5.    With  the  triangle  placed  as  in  sect.  47,  what 
will  be  the  height  of  the  tree  if  the  eye  is  4  ft.  6  in.  from 
the  ground  and  45  ft.  from  the  base  of  the  tree  ? 

5.  The  trunk  of  a  tree  when  trimmed  is  48  ft.  long.    At 
the  small  end  it  is  8  in.  in  diameter  and  at  the  large  end 
20  in.    If  we  cut  the  tree  into  four  12-ft.  lengths,  what  is 
the  largest  square  stick  of  timber  in  even  inches  that  can 
be  obtained  from  every  length  ? 

6.  If  we  saw  the  above  tree  into  ^-in.  boards,  allowing 
^-in.  waste  in  sawdust  for  every  cut,  how  many  feet  of 
lumber  shall  we  obtain  ? 

7.  The  height  of  a  tree  to  the  first  branch  is  40  ft.,  its 
diameter  at  the  base  is  26  in.,  and  at  the  branch  is  16  in. 
Allowing  2  in.  on  the  diameter  for  the  bark,  what  is  the 
largest  square  stick  of  timber  in  even  inches  that  can  be 
sawed  from  the  tree,  using  it  only  to  the  branch?    How 
many  feet  does  it  contain  ?    What  proportion  of  the  wood 
of  that  part  of  the  tree  used  does  the  timber  contain  ? 

8.  What  is  the  largest  square  stick  of  timber  in  even 
inches  that  can  be  sawed  from  a  log  16  in.  in  diameter  ? 
How  many  board  feet  does  it  contain  if  it  is  14  ft.  long  ? 


60  SHOP  PROBLEMS  IN  MATHEMATICS 

9.  How  many  1-in.  boards  can  be  sawed  from  the  above 
timber  if  ^  in.  is  allowed  for  sawdust  every  time  the  saw 
goes  through  ?    How  many  |-in.  boards  ?    How  many  £-in. 
boards  ? 

10.  If  a  California  sequoia  tree  measures  32  ft.  in  diame- 
ter at  the  base  and  2  ft.  in  diameter  300  ft.  from  the  ground, 
measurements  taken  inside  the  bark,  how  many  feet  of  1-in. 
boards  will  it  make,  deducting  3  in.  from  the  diameter  for 
slabs  and  £  for  sawdust  ?    (Use  the  average  diameter.) 

11.  How  many  rails  averaging  3  in.  square  and  10  ft. 
long  would  this  tree  make,  there  being  no  waste  in  slabs 
and  sawdust  ?    How  long  would  it  take  a  man  to  make  the 
rails  if  he  made  200  a  day  ? 

12.  If  a  stick  of  timber  is  16  in.  square,  how  far  must  we 
measure  each  way  from  the  corner  to  find  the  points  to  which 
to  chamfer  in  order  to  make  the  stick  a  perfect  octagon  ? 

13.  The  distance  between  any  two  parallel  sides  of  an 
octagonal  table  top  is  3  ft.    What  is  the  length  of  one  of 
its  sides  ? 

14.  On  a  piece  of  work  2  ft.  square  the  distance  to  be 
measured  each  way  from  a  corner  in  order  to  make  the 
work  octagonal  is  7  in.    How  far  from  the  corner  must 
we  measure  to  make  octagonal  a  piece  of  work  that  is 
17  in.  square? 

15.  The  top  of  a  derrick  for  hoisting  stone  is  32  ft.  high. 
How  many  feet  of  wire  rope  are  required  for  its  4  guy  ropes 
if  they  are  fastened  to  its  top  and  anchored  to  points  on 
the  ground  90,  100,  84,  and  124  ft.  respectively  from  the 
base  ?    (Add  12  ft.  for  fastenings.) 

16.  The  height  of  a  house  from  the  ground  to  the  eaves 
is  21  ft.    If  a  man  sets  the  foot  of  a  ladder  12  ft.  from  the 
house,  how  long  must  the  ladder  be  to  project  1  ft.  above 
the  eaves  ? 


„  a 


El 


CHAPTER  III 

PULLEYS, BELTS,  AND  SPEEDS 

PULLEYS  AND  SPEEDS 

55.  Driving  pulley.    The  driving  pulley  (Fig.  35)  is  the 
pulley  that  transmits  power  to  the  belt,  or,  in  other  words, 
causes  the  belt  to  move. 

56.  Driven   pulley.    The  driven  pulley  is  the  pulley  to 
which  the  power  is  transmitted  by  the  belt  or,  in  other 
words,  the  pulley  that  is  moved  by  the  belt. 

57.  Speed.    The  term  "  speed  "  as  applied  to  pulleys  means 
the  number  of  revolutions  they  make  per  minute,  and  is 
usually  designated  by  the  letters  K.  P.  M.  (revolutions  per 
minute). 

58.  Rule  for  speeds  of  pulleys.    The  diameter  of  the  driving 
pulley  multiplied  by  its  speed  is  equal  to  the  diameter  of  tit  e 
driven  pulley  multiplied  by  its  speed. 

59.  Cutting  speed.    The   term   "  cutting "   or   "  surface " 
speed  in  connection  with  a  lathe  means   the  number  of 
linear  feet  measured  on  the  surface  of  the  work  that  passes 
the  edge  of  a  cutting  tool  in  one  minute. 

60.  Rule  for  cutting  speed.    Multiply  the  number  of  feet  in 
the  circumference  of  the  work  being  turned  by  the  number  of 
revolutions  per  minute,  and  the  result  ivill  be  the  cutting  speed 
in  feet  per  minute. 

For  example,  if  the  circumference  of  a  piece  of  work 
that  is  being  turned  is  20  in.,  and  the  work  makes  400 
R.  P.  M.,  the  cutting  speed  is  400  x  f £  =  666|  ft.  per 
minute. 


PULLEYS,  BELTS,  AND  SPEEDS  63 


PROBLEMS 

1.  Let  c  equal  the  circumference  of  the  work  and  C  the 
cutting  speed.    Write  the  formula  for  C  in  terms  of  c  and 
E.  P.  M. ;  also  solve  for  c  and  K.'  P.  M. 

2.  Let  D  represent  the  diameter  of  the  driving  pulley, 
d  the  diameter  of  the  driven  pulley,  S  the  speed  of  the 
driving  pulley,  and  s  the  speed  of  the  driven  pulley. 

Write  the  formula  for  D  in  terms  of  d,  S,  and  s.    Solve 
for  d,  S,  and  s. 

3.  The  diameter  of  a  turned  cylinder  is  If  in.    What  is 
its  circumference? 

4.  What  is  the  diameter  of  a  turned  cylinder  the  cir- 
cumference of  which  is  11  in.? 

5.  Find  the  circumference  of  a  pulley  whose  radius  is 
10J  in.  ? 

6.  The  circumference  of  a  pulley  is  66^  in.    What  is  its 
radius  ? 

7.  The  diameter  of  a  driving  pulley  is  12  in.  and  its  speed 
is  400  R.  P.  M.    What  is  the  speed  of  the  driven  pulley 
whose  diameter  is  3  in.  ? 

8.  The  Pv.  P.  M.  of  the  cone  pulley  (see  Fig.  33)  on  a 
wood-turning  lathe  is  2500  and  the  diameter  of  one  step  is 
5  in.    What  must  be  the  diameter  of  the  driving  pulley 
if  its  R.  P.  M.  is  450? 

9.  The  driving  pulley  on  a  shaft  is  42  in.  in  diameter 
and  makes  20  revolutions  per  minute.    How  many  revolu- 
tions will  the  driven  pulley  make  if  its  diameter  is  3  ft.  ? 

10.  The  diameter  of  the  driving  pulley  is  9  in.  and  its 
speed  is  1000  R.  P.  M.    At  what   speed  will  the  driven 
pulley  run  if  its  diameter  is  4  in.? 


64  SHOP  PROBLEMS  IN  MATHEMATICS 

11.  If  the  speed  of  an  18-in.  pulley  on  the  driving  shaft 
is  300  E.  P.  M.,  what  is  the  speed  of  the  driven  shaft  which 
has  a  12-in.  pulley  belted  to  the  driver  ? 

12.  The  diameter  of  a  piece  that  is  being  turned  is  2  in. 
If  the  piece  makes  2500  revolutions  per  minute,  what  is  its 
cutting  speed  ? 

13.  The   surface  speed  of  a  turning  piece  of  work  is 
found  to  be  3000  ft.  per  minute.    If  its  diameter  is  4  in., 
what  is  its  E.  P.  M.  ? 

14.  What  is  the  diameter  of  a  cylinder,  if  its  E.  P.  M. 
is  2500  and  its  surface  speed  1500  ? 

15.  Sometimes  we  increase  the  speed  of  the  lathe  while 
turning  the  same  piece  of  work.    Explain  why  the  wood 
cuts  off  faster  when  we  do  this. 

16.  Using  the  same  speed,  we  turn  a  piece  down  from 
7^  in.  to  2|  in.  in  diameter.    What  is  the  ratio  of  the  cut- 
ting speeds  of  the  two  diameters  ? 

17.  A  mechanic  is  turning  on  the  flat  side  of  a  disk  8  in. 
in  diameter;  his  turning  chisel  is  placed  2  in.  from  the 
center  point.    What  is  the  difference  in  the  rate  of  the  cut- 
ting or  surface  speed,  if  he  moves  his  chisel  f  in.  towards 
the  edge  of  the  disk,  and  what  are  the  two  cutting  speeds 
if  the  E.  P.  M.  is  1600  ? 

18.  If  a  bicycle  wheel  is  30  in.  in  diameter,  how  many 
times  will  it  go  around  in  running  a  mile  ? 

19.  A  locomotive  drive  wheel  makes  290  revolutions  in 
running  a  mile.    What  is  its  diameter  ? 

20.  The  main  driving  wheel  of  an  engine  is  12  ft.  6  in.  in 
diameter  and  revolves  at  96  E.  P.  M. ;  it  is  belted  to  a  48-in. 
pulley  on  the  main  line  shaft.   Find  the  speed  of  the  shaft. 

21.  A  36-in.  pulley  making  143  E.  P.  M.  is  belted  to  an- 
other making  396  E.  P.  M.    Find  the  diameter  of  the  latter. 


PULLEYS,  BELTS,  AND  SPEEDS        65 

22.  A   certain   grindstone   will   stand   a  surface  or  rim 
speed  of  800  ft.  per  minute.    At  how  many  R.  P.  M.  could 
it  run  if  its  diameter  is  4  ft.  8  in.? 

23.  The  cutting  speed  of  a  turning  chisel  when  held  at 
the  rim  of  a  revolving  disk  is  a.   What  is  the  cutting  speed 
of  the  tool  if  it  is  moved  1  in.  towards  the  center,  the 
diameter  of  the  disk  being  d  ? 

24.  The  R.  P.  M.  of  a  wooden  disk  that  is  being  turned 
is  2500  and  its  diameter  is  6  in.    To  what  speed  must  the 
lathe  be  changed  to  make  the  surface  speed  on  the  edge 
of  the  disk  the  same  when  the  diameter  is  turned  down 
from  6  to  3J  in.  ? 

25.  The  surface  or  rim  speed  of  a  grindstone  is  500  ft. 
per  minute  and  its  diameter  is  30  in.    If  the  line  shaft  to 
which  it  is  belted  makes   300  R.  P.  M.  and  the  driving 
pulley  on  this  shaft  is  3  in.  in  diameter,  what  must  be 
the  diameter  of  the  driven  pulley  on  the  grindstone  shaft? 

26.  An  emery  wheel  is  16  in.  in  diameter  and  its  surface 
speed  is  2000  ft.  per  minute.    What  must  be  the  diameter 
of  the  driving  pulley  on  the  line  shaft  if  the  R.  P.  M.  of 
the  shaft  is  250  and  the  diameter  of  the  pulley  on  the 
emery-wheel  shaft  is  4|  in.  ? 

27.  A  motor  pulley  is  8  in.  in  diameter  and  the  motor 
makes  1200  R.  P.  M.    If  this  speed  is  increased  to  1600 
R.  P.  M.,  what  must  be  the  size  of  the  pulley  in  order  to 
give  the  belt  the  same  speed  ? 

28.  The  splicing  of  a  belt  connecting  two  equal  pulleys 
travels  through  the  air  at  the  rate  of  2000  ft.  per  minute. 
At  what  speed  must  the  pulleys  run  if  they  are  20  in.  in 
diameter  ? 

29.  The  surface  speed  of  an  emery  wheel  is  3000  ft.  per 
minute  and  its  R.P.  M.  is  600.    What  is  its  diameter  ? 


66 


SHOP  PROBLEMS  IN  MATHEMATICS 


30.  A  band  saw  (Fig.  34)  runs  over  two  wheels,  each  32 
in.  in  diameter.    If  the  band  saw  is  16  ft.  long  and  the 
speed  of  the  wheels  is  600  R.  P.  M.,  what  is  the  cutting 
speed  of  the  band  saw  ? 

31.  A  circular  saw  (Fig.  35)  makes  3000  R.  P.  M.  and  its 
diameter  is  16  in.    If  the  speed  of  the  upper  and  lower 
wheels  of  a  band  saw  is  450  R.  P.  M.  and  the  length  of 


Wheel 


Guard 


Star  fi  09 


FIG.  34.    BAND  SAW 
(By  permission  of  The  Crescent  Machine  Co.) 

the  saw  18  ft.,  what  must  be  the  diameter  of  the  band-saw 
wheels  in  order  that  its  cutting  speed  may  be  one  third  that 
of  the  circular  saw  ? 

32.  A  table  top  30  in.  in  diameter  is  being  turned.  What 
must  its  R.  P.  M.  be  if  its  surface  speed  at  the  outer  edge 
is  equal  to  that  of  a  cylinder  whose  diameter  is  2^  in.  and 
whose  R.  P.  M.  is  2500  ? 


PULLEYS,  BELTS,  AND  SPEEDS        67 

33.  The  front  wheel  of  a  carriage  is  3  ft.  4^-  in.  in  diam- 
eter.   How  many  turns  does  it  make  in  traveling  a  distance 
of  5£  mi.  ? 

34.  Two  boys  sitting  side  by  side  on  wooden  horses  ride 
on  a  merry-go-round.    The  distance  from  the  outer  horse 
to  the  center  is  16  ft.  and  the  distance  between  the  horses 
2  ft.    If  the  merry-go-round  makes  30  turns,  which  boy 
rides  farther,  and  how  much  ? 


Fence 


Driving 
Pulle 


FIG.  35.    CIRCULAR  SAW  TABLE 

(By  permission  of  The  Crescent  Machine  Co.) 

35.  How  many  revolutions   per  minute  are  made  by  a 
locomotive  driving  wheel  that  is  6  ft.  in  diameter,  if  the 
train  is  running  40  mi.  an  hour  ? 

36.  If  a  locomotive  wheel  4  ft.  in  diameter  goes  around 
9000  times  in  running  20  mi.,  how  much  distance  is  lost 
by  slipping  ? 


68  SHOP  PROBLEMS  IN  MATHEMATICS 

LENGTHS  OF  BELTS 

61.  Practical  measurement  of  belts.    Although  in  actual 
practice  the  length  of  a  belt  is  obtained  by  measuring  with 
a  tapeline,  or  otherwise,  after  the  pulleys  are  in  place,  a 
knowledge  of  how  the  length  is  obtained  mathematically 
is  of  no  small  value  in  the  study  of  machinery. 

62.  Rule  for  lengths  of  belts.    To  find  the  length  of  a  belt 
divide  the  sum  of  the  diameters  of  the  two  pulleys  by  2  and 
multiply  the  quotient  by  3=;  to  this  add  twice  the  distance 


a 


I 

FIG.  36.    PULLEYS  OF  EQUAL  DIAMETERS 
If  I  is  the  length  of  the  belt,  show  that  I  =7rD  +  2«. 

between  the  center  of  the  two  pulleys  and  the  result  will  be 
the  length  of  the  belt. 

This  rule  applies  only  to  the  cases  of  open  belts  in  Fig.  36  and 
Fig.  37.    Belts  are  open  unless  cross  belt  is  stated. 

63.  Degree  of  accuracy.    The  above  method  is  absolutely 
correct  only  when  the  pulleys  are  equal  in  diameter  (Fig. 
36).    The  error  obtained,  however,  when  the  pulleys  are  of 
different  diameters  (Fig.  37)  (unless  there  is  too  great  a 
difference)  is  so  slight  that  it  can  be  neglected,  for  all 
practical  purposes,  when  dealing  with  ordinary  leather, 
rubber,  or  canvas  belts. 

64,  Exact  lengths  of  belts.    If  it  is  desired  to  find  the 
exact  length  of  belts  when  the  pulleys  are  of  unequal  diam- 
eters, it  can  be  done  by  trigonometry  (see  sect.  249). 


PULLEYS,  BELTS,  AND  SPEEDS  69 

PROBLEMS 

1.  The  distance  between  the  centers  of  two  pulleys  of 
equal  diameters  (Fig.  36)  is  12  ft.  and  their  diameter  is 
21  in.    How  long  must  the  connecting  belt  be  ? 

2.  Each  of  two  pulleys  has  a  diameter  of  24  in.    If  the 
connecting  belt  is  28|-  ft.  long,  what  is  the  distance  between 
the  centers  of  the  pulleys  ? 

3.  If  the  distance  between  the  centers  of  two  pulleys  of 
equal  diameters  is  11^  ft.,  and  the  length  of  the  connecting 
belt  is  30  ft.,  what  is  the  diameter  of  the  pulleys  ? 


a 


FIG.  37.    PULLEYS  OF  UNEQUAL  DIAMETERS  WHERE 
THE  DIFFERENCE  is  SLIGHT 


Show  that    I  =  TT  --  I-  2  a. 

4.  How  much  belting  is  required  to  connect  two  pulleys 
24  and  18  in.  in  diameter  respectively  (Fig.  37),  the  dis- 
tance between  centers  being  9J  ft.? 

5.  If  the  diameter  of  the  driving  pulley  on  an  engine  is 
56  in.,  the  distance  between  its  center  and  the  center  of  a 
driven  pulley  on  a  line  shaft  is  18  ft.,  and  the  length  of 
the  connecting  belt  is  48  ft.,  what  is  the  diameter  of  the 
driven  pulley  ? 

6.  A  motor  is  belted  to  a  pulley  on  a  circular-saw  bench. 
If  the  length  of  the  belt  is  17  ft.  and  the  diameters  of  the 


70  SHOP  PROBLEMS  IN  MATHEMATICS 

two  pulleys  are  10  and  4  in.  respectively,  what  is  the  dis- 
tance between  the  centers  of  the  pulleys  ? 

7.  The  cone  pulley  on  a  lathe  has  three  steps  with  3,  5, 
and  7-in.  diameters  respectively,  and  the  corresponding 
cone  pulley  on  the  shaft  over  the  lathe  has  its  largest  step 
14  in.  in  diameter.  What  must  be  the  diameter  of  the 
other  two  steps  on  the  larger  pulley  in  order  to  have  the 
same  belt  work  on  all  three  steps  of  both  pulleys  ? 

J 


I 

FIG.  38.    PULLEYS  OF  UNEQUAL  DIAMETERS  WHERE 
THE  DIFFERENCE  is  GREAT 


Showthat    1  =  2 A/I'——1)  +  a2H  TT -• 


8.  A  driving  pulley  on  a  motor  is  5  in.  in  diameter  and 
the  driven  pulley  on  the  line  shaft  is  36  in.  in  diameter. 
If  the  distance  between  the  centers  is  8  ft.,  how  long  must 
the  connecting  belt  be  ? 

9.  Two  pulleys  are  belted  together  to  run  in  opposite 
directions  (cross  belt,  Fig.  39)  ;  the  diameter  of  each  is 
16  in.,  and  their  distance  between  centers  is  10  ft.    What 
length  of  belt  is  required  ? 

10.  Find  the  diameter  of  the  driving  pulley  on  a  motor,  if 
the  distance  between  its  center  and  the  center  of  the  driven 
pulley  on  the  lathe  is  22  in.,  the  length  of  the  connecting 
belt  is  60  in.,  and  the  diameter  of  the  driven  pulley  is  3  in. 


PULLEYS,  BELTS,  AND  SPEEDS        71 

11.  Two  pulleys  run  with  cross  belt  (Mg.  40).  The  diam- 
eter of  one  is  30  in.  and  the  diameter  of  the  other  is  16  in. 
What  length  of  belt  do  they  require  if  the  distance  between 
centers  is  13^  ft.  ? 

a         A 


i  I 

Fio.  S..).    PULLEYS  OF  EQUAL  DIAMETERS  WITH 
CROSS  BELT 

Show  that    I  =  2  VZ>2  -f  a2  -f  irD. 

12.  Two  pulleys  of  equal  diameters  are  connected  with 
a  cross  belt.    If  the  length  of  this  belt  is  29  ft.  and  the 


FIG.  40.    PULLEYS  OF  UNEQUAL  DIAMETERS  WITH 
CROSS  BELT 


Show  that    I  =  2\l[  — )  +  a2  +  TT  — 


distance  between  centers  of  the  pulleys  is  7  ft.  5  in.,  what 
is  the  diameter  of  the  pulleys  ? 

13.  The  diameters  of  two  pulleys  are  to  each  other  as  3 
is  to  5,  and  the  distance  between  their  centers  is  8  ft.    If 


72  SHOP  PROBLEMS  IN  MATHEMATICS 

the  diameter  of  the  larger  pulley  is  25  in.,  what  is  the 
length  of  the  connecting  belt? 

14.  What  is  the  length  of  the  connecting  belt  if  the  dis- 
tance between  the  two  pulleys  is  15  ft.  and  the  diameters 
of  the  pulleys  are  24  in.  and  8  in.  respectively  ? 

15.  The  diameters  of  two  pulleys  are  to  each  other  as  9 
is  to  10.    If  the  diameter  of  the  smaller  pulley  is  18  in.  and 
the  distance   between  the  centers   is  19  ft.,  what  is  the 
length  of  the  connecting  belt  ? 

16.  The  diameters  of  two  pulleys  are  16  in.  and  18  in. 
respectively,  and  the  distance  between  their  centers  is  11^  ft. 
If  the  belt  is  8  in.  wide,  how  many  square  feet  of  belting  are 
required  ?  How  many  square  feet  if  the  belt  is  6  in.  wide  ? 

17.  The  distance  between  centers  of  two  pulleys  is  12  ft. 
and  the  smaller  of  the  pulleys  is  16  in.  in  diameter.    What 
is  the  length  of  the  belt  if  the  diameters  of  the  pulleys  are 
to  each  other  as  2  is  to  3  ? 

18.  How  much  leather  (surface  measure)  will  it  take  for 
a  belt  6  in.  wide  to  connect  two  pulleys  whose  diameters 
are  16  in.  and  20  in.  respectively,  the  distance  between  cen- 
ters being  9£  ft.  ? 

19.  At  $1  per  square  foot,   what  will  be  the  cost  of 
leather  belting  for  the  following : 

Belt  to  be  crossed  and  5  in.  wide. 
Distance  between  centers,  15  ft.  4  in. 
Small  pulley,  18  in.  diameter. 
Diameters  of  pulleys  in  the  ratio  5  : 6. 

COILS  OF  BELTING,  ETC. 

65.  Rule  for  belting.  To  find  the  number  of  feet  of  belting 
in  a  roll,  add  the  diameter  of  the  roll  to  the  diameter  of  the 
hole  left  in  the  center,  divide  the  sum  by  2,  multiply  by  y, 
and  multiply  this  result  by  the  number  of  coils  in  the  roll. 


PULLEYS,  BELTS,  AND  SPEEDS        73 

EXAMPLE.  A  roll  of  belting  contains  60  coils  ;  its  outside 
diameter  is  36  in.  and  the  diameter  of  the  hole  in  the  center 
is  6  in.  How  many  feet  does  the  roll  contain  ? 

op      I       f  OO  OO 

22LT.2  x  --  x  60  =  21  x  -^  X  60  =  3960 in.  =  330  ft. 


PROBLEMS 

1.  At  $1  per  linear  foot,  what  will  be  the  cost  of  a  roll 
of  belting  whose  outside  and  inside  diameters  are  30  and 

5  in.  respectively,  the  number  of  coils  being  40  ? 

2.  The  length  of  a  piece  of  belting  is  80  ft.,  its  width  is 

6  in.,  and  its  thickness  T\  in.    Find  how  many  coils  it  will 
make  and  what  will  be  the  diameter  of  the  roll  if  the  hole 
left  in  the  center  is  4  in.  in  diameter  ? 

3.  A  cable  ^  in.  in  diameter  is  wrapped  around  a  roller 
12  in.  in  diameter.    How  many  coils  will  be  required  to 
reach  the  bottom  of  a  shaft  120  ft.  deep  ? 

4.  A  hoisting  windlass  drum  or  roller  is  18  in.  in  diameter 
on  one  end  and  9  in.  in  diameter  on  the  other.    A  rope  has 
one  end  winding  on  the  18-in.  end  of  the  drum.    When  it 
has  passed  around  a  tackle-block  pulley  below,  it  runs  up 
and  winds  around  the  9-in.  end  of  the  drum  in  the  opposite 
direction.   If  the  windlass  makes  25  turns  per  minute,  how 
fast  does  the  load  fastened  to  the  tackle  block  rise  ? 


CHAPTER   IV 

AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS  ; 
TURNED  WORK 

SHEET-METAL  WORK 

66.  Formulas  for  sheet-metal  work.  For  formulas  of  sur- 
face and  cubic  measure  that  apply  to  problems  in  sheet- 
metal  work  the  student  is  referred  to  Chapter  XVI. 


PROBLEMS 

1.  A  galvanized-iron  measure  (Fig.  41)  14  in.  in  diameter 
and  14  in.  deep  holds  very  nearly  a  bushel.  Allowing  2  in. 

on  the  width  of  the  side  piece 

^- — • -^^      for  the  lock  at  the  bottom  and 

the  roll  at  the  top,  1  in.  on  the 
length  for  the  side  seam  (lock), 
and  1  in.  on  the  diameter  of  the 
bottom  for  the  lock,  how  much 
metal  is  required  for  its  con- 
struction ? 

2.  If  the  above  measure  were 
16  in.  in  diameter,  how  deep 
would  it  have  to  be  to  hold 
a  bushel  ?  (A  bushel  contains 
2150  cu.  in.) 

3.  If  the  measure  mentioned  in  Problem  1  were  9^  in. 
deep,  what  would  its  diameter  have  to  be  to  hold  a  bushel 
and  a  half  ? 

74 


FIG.  41.    BUSHEL  MEASURE 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      75 

4.  If  we  make  the  depth  of  the  above  measure  20  in., 
what   must  be  its  diameter  to  hold  a  bushel?    Would  it 
contain  more  or  less  metal,  and  how  much  ? 

5.  A  copper  clothes  boiler   has   semicircular  ends  ;    its 
extreme  length  is  21  in.,  its  width  11^  in.,  and  its  depth 
13  in.    If  we  allow  1^  in.  on  the  width  of  the  side  piece 
for  lock  and  top  roll,  1  in.  on  the  length  for  the  side  lock, 
and  ^  in.  all  around  the  bottom  piece  for  the  lock,  how 
much  copper  is  required  to  construct  the  boiler  ? 

6.  If  the  boiler  above  were  11  in.  wide  and  21^  in.  long 
and  held  the  same  amount,   what  would  be  its   depth  ? 
Would  it  contain  more  or  less 

copper,  and  how  much  ? 

7.  A    galvanized-iron    water 
pail  is  10^-  in.  in  diameter  at 
the  top,  8^  in.  at  the  bottom, 
and  9^  in.  in  slant  height.    If 

we  allow  14^  in.  extra  for  bottom 

FIG.  42.    CIRCULAR  HOPPER 

lock  and  top  roll,  1  in.  for  side 

lock,  and  1  in.  on  the  diameter  of  the  bottom  piece  for  lock, 

how  much  metal  is  required  for  its  construction  ? 

8.  A  tin  basin  is  16  in.  in  diameter  at  the  top  and  14  in. 
in  diameter  at  the  bottom.    How  much  tin  is  required  for 
its  construction  if  its  slant  height  is  4^  in.  ?    Allow  1  in. 
extra  on  the  length  and  width  of  two  pieces  for.  the  side 
and  1  in.  on  the  diameter  of  the  bottom  piece  for  locks. 

9.  How  much  tin  is  required  for  the  above  basin  if  its 
perpendicular  height  or  depth  is  3%  in.? 

10.  A  circular  hopper  (Fig.  42)  is  constructed  of  two 
pieces  of  galvanized  iron  ;  its  perpendicular  height  is  20  in., 
its  large  diameter  is  40  in.,  and  its  small  diameter  is  3^  in. 
How  much  metal  is  required  for  its  construction  if  1  in.  is 
added  to  each  piece  for  lock  ? 


76 


SHOP  PROBLEMS  IN  MATHEMATICS 


11.  A  funnel  (Fig.  43)  is  6  in.  in  diameter  at  the  top 
and  1  in.  at  the  bottom,  its  slant  height  is  4|-  in.,  the  spout 
is  1  in.  in  diameter  at  the  larger  end  and  \  in.  at  the 
smaller,  and  its  slant  height  is  3J  in.  Find  the  amount  of 
tin  required  if  we  allow  ^  in.  on 
the  length  and  width  of  each  piece 
for  locks.  (See  Fig.  44  for  laying 
out  pattern  for  funnel.) 

12.  A  cone-shaped  hood  for  a 
chimney  is  a  in.  in  diameter  at 
the  bottom  and  has  a  slant  height 
of  b  in.  Find  the  amount  of  metal 


FIG.  43.    FUNNEL 


FIG.  44.    PATTERN  FOR  FUNNEL 


in  the  hood  if  1  in.  is  added  to  the  length  of  the  piece  for 
a  side  lock. 

13.  If  the  perpendicular  height-  of  the  above  hood  is 
11  in.  and  a  equals  28  in.,  how  much  metal  is  required  for 
the  hood,  allowing  1  in.  on  the  side  for  lock  ? 

14.  If  a  sheet  of  tin  is  20  by  28  in.,  how  many  disks  for 
can  bottoms  3f  in.  in  diameter  will  it  make,  and  how  much 
tin  will  be  wasted  ? 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      77 

15.  How   much   tin    (surface   measure)   is   required   for 
1  gross  of  tin  cans  3f  in.  in  diameter  and  4|  in.  high,  allow- 
ing f  in.  on  the  diameter  of  the  end  pieces,  ^  in.  on  the 
length,  and  1  in.  on  the  height  of  the  side  pieces  for  locks  ? 

16.  If  iron  ^  in.  thick  weighs  5  Ib.  per  square  foot,  what 
will  be  the  weight  of  iron  for  the  following  tank  :  extreme 
length,  4  ft. ;  width,  2  ft. ;  depth,  2  ft. ;  ends  to  be  true 
semicircles,  2  in.  extra  allowed   on   the  length  and  1  in. 
extra  on  the  width  of  the  side  piece,  and  1  in.  extra  all 
around  the  edge  of  the  bottom  piece  for  locks  (seams)  ? 

17.  The  size  of  sheets  of  tin  is  20  by  28  in.    If  we  allow 
1  in.  on  both  the  length  and  width  of  each  sheet  for  lock, 
how  much  tin  (how  many  sheets)  is  required  to  cover  a 
roof  36  ft.  wide  and  50  ft.  long  ?  At  5  cents  per  square  foot, 
what  is  the  cost  of  the  tin  ? 

18.  A  certain  make  of  tin  shingles  9  by  14  in.  when  laid 
is  exposed  to  the  weather  8  in.  by  12  in.    At  $6  per  square 
(100  sq.  ft.),  what  will  be  the  cost  of  covering  a  roof  26  ft. 
wide  and  34  ft.  long  ?    How  many  shingles  willit  take  ? 

19.  If  sheet  lead  T^  in.  thick  weighs  3.69  Ib.  to  the  square 
foot,  how  many  pounds  will  be  required  to  line  the  four 
sides  and  bottom  of  a  tank  3  ft.  6  in.  deep,  4  ft.  wide,  and 
8  feet  long  ? 

20.  Copper  ^  in.  thick  weighs  2.888  Ib.  per  square  foot. 
A  man  cuts  from  a  sheet  of  this  copper  3  equal  disks,  each 
tangent  to  all  the  others  and  12  in.  in  diameter.  What  is  the 
weight  of  the  piece  left  in  the  center  after  the  disks  are 
cut  out  ? 

21.  If  number  20  copper  weighs  1.5855  Ib.  per  square 
foot,  what  will  be  the  weight  of  enough  metal  to  line  a 
range  reservoir  that  is  30  in.  long,  10  in.  wide,  and  11  in. 
deep,  allowing  J  sq.  ft.  for  locks  and  waste  ? 


78  SHOP  PROBLEMS  IN  MATHEMATICS 

22.  A  copper  teapot  is  9f  in.  in  diameter  at  the  bottom, 
7  in.  at  the  top,  and  10  in.  deep.  Allowing  40  sq.  in.  for 
locks  and  waste,  how  much  metal  is  required  for  its  con- 
struction without  a  cover  ? 


TANKS,  RESERVOIRS,  CISTERN^,  ETC. 

67.  Exact  measure.  To  find  the  exact  cubic  contents  of 
any  containing  vessel,  the  student  is  referred  to  Chap- 
ter XVI. 

PROBLEMS 

1.  If  a  can  is  5  in.  square  on  the  bottom,  how  high  must 
it  be  to  hold  a  gallon  ?    (A  gallon  contains  231  cu.  in.) 

2.  A  reservoir  tank  on  a  range  is  20  in.  long  and  10  in. 
wide.    How  deep  must  it  be  to  hold  10  gal.  ? 

3.  How  much  sheet  lead  is  required  to  line  a  tank  5  ft. 
square  on  the  bottom,  the  tank  to  hold  300  gal.  ? 

4.  A  tank  1  ft.  square  on  the  bottom  and  18  in.  high 
holds  how  many  gallons  ? 

5.  A  vessel  7  in.  in  diameter  and  6  in.  deep  holds  how 
much  ? 

6.  A  can  that  is  10  in.  in  diameter  must  be  how  deep  to 
hold  5  gal.  ? 

7.  If  a  milk  can  is  24^-  in.  high,  what  must  be  its  diame- 
ter to  hold  40  qt.  ? 

8.  An  ash  can  is  18  in.  in  diameter  and  30  in.  high.    How 
many  bushels  does  it  hold  ?   (A  bushel  contains  2150  cu.  in.) 

9.  If  the  depth  of  a  galvanized  measure  is  12  in.,  what 
must  be  its  diameter  to  hold  a  bushel  ? 

10.  What  must  be  the  diameter  of  a  vessel  that  is  8  in. 
deep,  to  hold  a  bushel  ?  to  hold  a  half  bushel  ?   to  hold  a 
peck  ? 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      79 

11.  A  boy  lets  a  string  with  a  weight  on  its  end  down 
into  a  cistern  that  is  circular  in  shape  ;  when  the  weight 
strikes  the  bottom  and  the  string  is  drawn  up  he  finds  it 
wet  3  ft.  and  5  in.    If  the  cistern  is  5  ft.  in  diameter,  how 
many  gallons  of  water  does  it  contain  ? 

12.  A  vertical  water  pipe  is  2  in.  in  diameter,  inside  meas- 
ure.   What  is  the  pressure  per  square  inch  on  the  bottom 
end  of  the  pipe,  if  water  weighs  62.5  Ib.  per  cubic  foot 
and  the  length  of  the  pipe  is  33  ft.  ? 

13.  A  standpipe  is  40  ft.  high  and  16  ft.  in  diameter, 
inside  measure.    What  is  the  pressure  in  pounds  on  the 
bottom  of  the  pipe  when  it  is  f  full  ? 

14.  What  must  be  the  diameter  of  a  cistern  7  ft.  deep 
to  hold  35  bbl.  ?    (A  barrel  contains  31^  gal.) 

15.  A  copper  tank  is  11^  in.  wide,  13  in.  deep,  and  21  in. 
long.   If  the  ends  are  semicircular  in  form,  how  many  gallons 
does  it  hold  ? 

16.  A  silo  is  16  ft.  in  diameter  and  30  ft.  high.    How 
much  matched  lumber  J-  in.  thick  is  required  to  cover  its 
outside,  allowing  50  ft.  for  waste  in  squaring  off  the  ends 
of  the  boards  ? 

17.  If  a  water  boiler  is  12  in.  in  diameter,  how  high  must 
it  be  to  hold  30  gal.  ? 

18.  If  a  water  boiler  is  5  ft.  high,  what  must  be  its 
diameter  to  hold  52  gal.? 

19.  A  water  boiler  is  22  in.  in  diameter  and  holds  100 
gal.    What  is  its  height  ? 

20.  If  the  boiler  mentioned  in  Problem  19  is  made  of 
TVin.  iron,  what  will  be  its  weight  if  we  add  1  in.  for  the 
lap  joint  (lock)  on  the  side  and  2£  in.  to  the  diameter  of 
the  end  pieces  for  the  same  purpose,  jVin.  iron  weighing 
12.673  Ib.  per  square  foot  and  the  rivets  10  Ib.  extra  ? 


80  SHOP  PROBLEMS  IN  MATHEMATICS 

21.  A  flat  tin  roof  is  20  by  30  ft.  and  a  conductor  pipe 
leads  from  the  roof  to  a  cistern  8  ft.  in  diameter.    When  it 
began  to  rain  the  cistern  was  empty,  and  there  were  4  ft. 
of  water  in  it  when  the  rain  stopped.    How  many  inches 
of  rain  fell  ? 

22.  If  the  above  were  a  common  gable  roof  25  ft.  long, 
with  19J-ft.  rafters  having  a  half  pitch,  what  would  be  the 
number  of  inches  of  rainfall  necessary  to  produce  the  same 
amount  of  water  in  the  cistern  ? 

23.  If  a  ^-pitch  roof  is  31  ft.  long  and  has  rafters  19  ft. 
6  in.  long,  how  many,  gallons  of  water  will  run  into  the 
cistern  if  the  rainfall  is  £  in.  ? 

24.  A  railroad  water  tank  is  16  ft.  in  diameter  and  16  ft. 
high.   How  many  gallons  does  it  hold  ?  how  many  barrels? 
(31*  gal.) 

68.  Formulas  for  areas  and  volumes.  Before  working  the 
problems  in  this  chapter  the  student  should  read  Chapter 
XVI. 

PROBLEMS 

1.  A  cylinder  3  in.  in  diameter  and  6  in.  long  is  being 
turned  from  a  block  3£  in.  square  and  7  in.  long.    How 
much  wood  is  cut  away  in  shavings  ? 

2.  A  manual-training  class  is  composed  of  30  boys,  and 
every  boy  turns  up  a  cylinder  If  in.  in  diameter  and  6  in. 
long  from  a  piece  of  rough  stock  2  in.  square  and  7  in.  long. 
At  $70  per  M,  how  much  will  the  lumber  cost  and  how 
much  of  the  wood  is  actually  used  in  the  cylinders  ? 

3.  A  disk  5  in.  in  diameter  and  1  in.  thick  is  turned  from 
a  piece  of  stock  5J  in.  square  and  1£  in.  thick.    How  much 
of  the  wood  is  wasted  in  shavings  ? 

4.  A  croquet  ball  is  3f  in.  in  diameter  and  the  mallet 
heads  are  2£  in.  in  diameter  and  5f  in.  long.    If  we  allow 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      81 


f  in.  on  the  length  and  J  in.  on  the  other  two  dimensions 
of  every  piece  for  rough  stock,  how  many  feet  of  hard  maple 
lumber  is  required  for  a  set  of  8  balls  and  mallet  heads  ? 
How  much  of  the  lumber  is  wasted  in  shavings  ? 

5.  What  will  be  the  weight  of  a  bowling  ball  6  in.  in  di- 
ameter that  is  to  be  turned  from  a  rectangular  block  6^  in. 
square  and  7  in.  long,  the  weight  of  the  block  being  13J  Ib.  ? 

6.  A  lignum- vitse  bowling  ball  is  8  in.  in  diameter  and 
weighs  10  Ib.    What  was  the 

weight  of  the  cube  from  which 
it  was  turned,  if  we  allow  J  in. 
on  each  of  the  three  dimensions 
for  rough  stock  ? 

7.  A  dumb-bell  is  turned  from 
a  rectangular  piece  of  wood  3^ 
by  3£  by  12  in.    The  balls  on 
the  end  are  3  in.  in  diameter, 
and  are  connected  by  a  handle 
1^  in.  in  diameter  and  4  in.  long. 
Find  how  much  wood  is  cut  away 
in  shavings,  and  how  much  the 
dumb-bell  weighs,  if  the  piece  be- 
fore being  turned  weighed  5-J-  Ib. 

8.  What  is  the  weight  of  a  cast-iron  sphere  5  in.  in  diam- 
eter, if  cast  iron  weighs  .26  Ib.  per  cubic  inch  ? 

9.  How  many  cubic  inches  in  a  hollow  cast-iron  sphere 
6  in.  outside  and  4  in.,  inside  diameter  ? 

10.  What  is  the  weight  of  the  above  sphere,  if  the  specific 
gravity  of  cast  iron  is  7.202  ? 

11.  At  n  cents  per  pound,  what  will  be  the  cost  of  a 
brass  sphere  d  inches  in  diameter  ? 

12.  A  ring  (Fig.  45)  has  a  cross  section  1  in.  square  and  an 
outside  diameter  of  6  in.    What  is  the  volume  of  the  ring  ? 


FIG.  45.    RING  WITH  SQUARE 
CROSS  SECTION 


82  SHOP  PROBLEMS  IN  MATHEMATICS 

13.  The  cross  section  of  the  rim  of  a  flywheel  is  2^  by 
3£  in.    If  its  outside  diameter  is  30  in.  and  the  inside  diam- 
eter 23  in.,  what  is  the  volume  of  the  rim  ? 

14.  In  Problem  13  the  inside  diameter  is  23  in.,  the  width 
of  the  riin  2^  in.,  and  its  depth  3^  in.    If  the  diameter  were 
the  same  and  the  rim  were  built  the  other  way,  with  the 
width  3^-  in.  and  the  depth  2^  in.,  would  there  be  more  or 
less  metal  ?   How  much  ? 

15.  The  weight  of  a  cast-iron  ring  with  a  square  cross 
section  is  200  Ib.  and  its  mean  diameter  is  35  in.    Find 

the  outer  diameter  of  the  ring 
and  its  volume. 

16.  The  outer  and  inner  diam- 
eters of  the  rim  of  a  cast-iron 
handwheel  are  13^  and   12   in. 
respectively.  What  is  the  weight 
of  the  rim  if  its  cross  section  is 
a  circle  ?    (Fig.  46.) 

17.  The  area  of  the  circular 

cross    section    of   the   rim   of   a 
FIG.  46.  RING  WITH  CIRCULAR  .  _       .       .  .     „_„   .  - 

CROSS  SECTION  cast-iron  flywheel  is  20f  in.  and 

its  weight  is  300  Ib.    What  are 
its  inner  and  outer  diameters  ? 

18.  The  rim  of  a  cast-iron  handwheel  has  an  octagonal 
cross  section  f  in.  on  a  side.    If  its  volume  is  65  cu.  in., 
what  are  its  inner  and  outer  diameters  ? 

19.  A  brass  ring  having  an  inner  diameter  of  4  in.  and  a 
cross-section  diameter  of  1  in.  is  melted  down  and  run  into 
a  ball.    What  is  the  diameter  of  the  ball  ? 

20.  A  brass  ring  1  sq.  in.  in  cross  section  and  5  in.  out- 
side diameter  is  melted  and  run  into  a  ring  with  a  1-in. 
cross-section  diameter  and  the  same  outside  diameter.    How 
much  brass  will  be  left  ? 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      83 

PROBLEMS 

TANKS 

1.  A  wooden  tank  is  6  ft.  square  at  the  bottom  and  5  ft. 
square  at  the  top.    If  it  is  4  ft.  6  in.  deep,  how  many  gallons 
does  it  hold  ? 

2.  A  wooden  tank  is  6  ft.  5  in.  in  diameter  at  the  top, 
6  ft.  10  in.  in  diameter  at  the  bottom,  and  4  ft.  deep,  inside 
measure.    How  many  barrels  does  it  contain  ?    (A  barrel 
contains  31J  gal.) 

3.  If  a  wooden  churn  holds  7  gal.,  is  20  in.  deep,  and 
10  in.  in  diameter  at  the  bottom,  what  must  its  diameter 
be  at  the  top  ? 

4.  If  the  above  churn  was  24  in.  deep,  12  in.  in  diameter 
at  the  bottom,  and  9  in.  in  diameter  at  the  top,  how  many 
gallons  would  it  hold? 

5.  What  would  be  the  depth  of  a  churn  that  would  hold 
15  gal.,  if  its  top  and  bottom  diameters  were  10  in.  and 
14  in.  respectively  ? 

6.  The  staves  of  a  water  tank  are  3f  in.  wide  at  the 
bottom  and  3^  in.  at  the  top.    If  the  tank  is  8  ft.  in  diam- 
eter at  the  bottom,  what  is  its  diameter  at  the  top  and  how 
many  staves  are  required  ? 

7.  A  water  tank  14  ft.  in  diameter  is  built  of  2-in.  by 
5-in.  staves  12  ft.  long  in  the  rough.    If  we  allow  £  in.  on 
the  width  of  every  stave  for  jointing,  how  many  staves  are 
required  and  how  much  lumber  do  they  contain  ? 

8.  How  much  paint  will  it  take  to  give  the  above  tank 
two  coats  all  over,  if  a  gallon  of  paint  will  cover  500  sq.  ft.  ? 

9.  A  cylindrical   flume    constructed    of    wooden    staves 
is  ^  mi.   long  and   3  ft.  in    outside  diameter.     Find  how 
many  staves  2  in.  by  4  in.  by  10  ft.  are  required  for  its 


84  SHOP  PROBLEMS  IN  MATHEMATICS 

construction,  if  we  allow  £  in.  on  the  width  of  each  stave 
for  jointing. 

10.  At  $35  per  M,  what  will  be  the  cost  for  lumber  for 
the  flume  mentioned  in  Problem  9  ? 

HEAPS 

11.  A  farmer  has  a  conical-shaped  heap  of  grain  on  his 
granary  floor  ;  the  height  of  the  heap  is  4  ft.  and  the  diam- 
eter of  the  circle  covered  by  its  base  is  10  ft.    If  he  puts 
the  grain  in  a  bin  4  ft.  square,  how  deep  will  it  be  and  how 
many  bushels  will  he  have  ? 

12.  An  amount  of  pea  coal  is  thrown  up  against  the  side 
of  a  bin.    The  highest  point  of  the  coal  is  6  ft.,  where  it 
touches  the  side  of  the  bin,  and  the  slant  height  of  the 
heap  is  9  ft.    If  the  base  of  the  heap  is  semicircular  in 
shape,  how  many  tons  does  it  contain,  allowing  36  cu.  ft. 
per  ton  ?    (This  holds  approximately  for  the  ordinary  sizes 
of  anthracite  coal.) 

13.  A  sack  of  grain  containing  2  bu.  is  emptied  out  on  a 
level  floor.    The  heap  naturally  takes  the  form  of  a  cone. 
If  its  highest  point  is  14  in.,  what  is  the  diameter  of  the 
circle  on  the  floor  covered  by  the  grain  ? 

14.  How  many  bushels  of  wheat  are  there  in  a  heap 
thrown  into  the  corner  of  a  bin,  if  the  highest  point  of 
the  corner  is  4  ft.  2  in.  and  the  distance  of  the  edge  of 
the  heap  from  the  corner  on  the  floor  is  6  ft.  6  in.? 

CONTENTS  OF  BARRELS,  CASKS,  ETC. 

69.  To  compute  the  contents  of  barrels,  casks,  etc.  There 
are  two  methods  for  computing  the  contents  of  barrels, 
casks,  etc.  The  first  is  approximate,  the  second  more  nearly 
exact. 


AREAS,  VOLUMES,  AND  WEIGHTS  OF  SOLIDS      85 

(1)  Add  the  smaller  and  the  greater  diameters  together 
(the  head  and  bung  diameters)  and  divide  by  2.   This  gives 
the  mean  diameter.    We  can  then  proceed  as  though  the 
barrel  were  a  common  c}dindrical  vessel  with  a  diameter 
equal  to  this  mean  diameter. 

(2)  Multiply  the  square  of  the  mean  diameter  in  inches 
by  .0034,  and  this  product  by  the  length  of  the  barrel  in 
inches.    The  result  will  be  the  number  of  gallons. 

70.  Degree  of  accuracy.   It  is  evident  that  the  more  curva- 
ture there  is  in  the  staves  of  a  barrel  the  greater  error 
there  will  be  in  using  the  first  method.    It  is  also  evident 
that  because  the  curvature  of  the  staves  in  some  barrels 
and  casks  is  greater  than  in  others,  no  general  rule  can  be 
formed  that  will  be  correct  in  all  cases.   The  second  method, 
however,  will  be  found  to  be  accurate  enough  for  all  practi- 
cal purposes. 

71.  Gallons  in  a  barrel  and  in  a  hogshead.  A  common  barrel 
is  said  to  contain  31^  gal.  and  a  hogshead  63  gal.    As  a 
matter  of  fact  they  are  made  in  a  number  of  sizes,  to  meet 
the  various  uses  to  which  they  are  put. 

The  dimensions  given  in  the  following  problems  are  inside  dimen- 
sions, unless  otherwise  stated. 

PROBLEMS 

1.  If  the  head  diameter  of  a  barrel  is  17^  in.,  its  bung 
diameter  20£  in.,  and  its  length  2^-  ft.,  how  many  gallons 
does  it  hold  ?    (By  first  method.) 

2.  A  barrel  has  a  bung  diameter  of  24  in.,  a  head  diameter 
of  20  in.,  and  is  30  in.  long.   How  many  gallons  will  it  hold? 
(By  second  method.) 

3.  A  barrel  holds  40  gal.    If  its  length  is  27.9  in.  and 
its  head  diameter  19  in.,  what  is  its  bung  diameter  ?    (By 
first  method.) 


86  SHOP  PROBLEMS  IN  MATHEMATICS 

4.  If  a  barrel  holds  61  gal.,  has  a  length  of  34  in.,  and 
a  bung  diameter  of  24£  in.,  what  is  its  head  diameter  ? 
(By  second  method.) 

5.  The   ordinary  sugar  barrel  is  about  19  in.  in  head 
diameter,  23  in.  in  bilge  diameter,  and  28  in.  long.    How 
many  of  these  barrels  will  it  take  to  hold  a  half  ton  of 
pea  coal  ?    (By  first  method.) 

6.  How  many  gallons  of  vinegar  in  a  cask  19  in.  in  head 
diameter,  23  in.  in  bung  diameter,  and  30  in.  long  ?    (By 
second  method.) 

7.  A  man  has  10  bbl.  of  potatoes  for  which  he  wishes 
to  build  a  bin.    The  barrels  are  18  in.  in  head  diameter, 
21  in.  in  bung  diameter,  and  30  in.  long.    If  he  builds  his 
bin  4  ft.  6  in.  by  5  ft.  4  in.,  how  deep  must  it  be  to  hold 
the  potatoes,  and  how  many  bushels  does  he  have  ?    (By 
first  method.) 

8.  A  dealer  delivers  3  bbl.  of  coal  ;  the  head  and  bung 
diameters  of  the  barrels  are  19  and  23  in.  respectively  and 
their  length  is  28  in.    How  much  ought  he  to  receive  for 
the  coal  if  he  charges  $7.50  per  ton  ?    (By  first  method.) 


CHAPTER  V 

PATTERN  MAKING  AND  FOUNDRY  WORK ;  WEIGHTS  OF 
CASTINGS  FROM  WEIGHTS  OF  PATTERNS 

72.  Weight  of  metal.    Foundrymen  usually  have  special 
rules  by  which  they  estimate  the  weights  of  castings  from 
the  weights  of  patterns  (see  table  of  weights).    When  the 
castings  have  small  openings,  their  cores  and  core  prints 
need  not  be  considered  in  weighing  the  patterns.    When, 
however,   the   castings   involve  large  dry  sand  cores,  the 
approximate  amount  of  metal  required  can  be  obtained  by 
multiplying  the  weight  of  the  dry  sand  core  by  the  proper 
factor  (see  table)  and  deducting  this  amount  from  the  weight 
obtained  for  the  casting  as  though  it  were  to  be  solid. 

It  will  readily  be  seen  that  the  above  rules  are  applicable  only  in 
cases  where  the  patterns  are  of  solid  wood  throughout.  As  most  of 
the  larger  patterns  are  constructed  more  or  less  hollow,  it  is  impossible 
to  formulate  any  general  rule  that  will  apply  to  all  cases. 

73.  Allowance  for  shrinkage  and  waste.   In  the  tables  given 
in  the  back  of  the  book  the  factors  by  which  to  multiply  are 
a  little  larger  than  figure  out  from  the  actual  weights  of  the 
various  woods  and  metals.    The  extra  amount  is  added  to 
allow  for  any  shrinkage  of  the  metal  in  melting,  waste  in 
sprues,  etc.,  or  variation  in  the  weight  of  wood  in  the 
patterns. 

PROBLEMS 

1.  A  solid  cast-iron  cylinder  is  3  in.  in  diameter.  What 
length  must  we  make  the  pattern  to  have  the  casting  weigh 
15  Ib.  ? 

87 


88 


SHOP  PROBLEMS  IN  MATHEMATICS 


2.  If  window  weights  are  1^  in.  in  diameter,  how  long 
must  we  make  the  pattern  for  7  Ib.  weights  ?    9  Ib.  ? 

3.  The  casting  for  a  solid  brass  roller  weighs  50  Ib.   What 
will  be  the  weight  of  a  cherry  pattern  for  the  roller  if  the 
specific  gravity  of  cherry  is  .715  and  that  of  brass  7.82  ? 


^Balancing  Point 

— 

Core  Print 

FIG.  47.    RETURN  BEND 

4.  A  boy  wishes  to  make  a  pattern  for  a  cast-iron  dumb- 
bell that  will  weigh  20  Ib.    If  the  handle  connecting  the 
two  balls  is  4  in.  long  and  1^  in.  in  diameter,  what  must  be 
the  size  of  the  balls  ? 

5.  The    mean    circumference    of    a    return-bend    pattern 
(Fig.  47)  is  c,  and  the  diameter  of  a  cross  section  of  the  open- 
ing is  2  r.    Find  I,  the  length  of  the  core  prints  required  for 
the  balanced  core,  the  diameter  of  which  is  d. 

Make  the  volumes  of  the  parts  of  the  core  outside  and  inside  of  the 
casting  equal. 

6.  A  saw-bench  top  is   made   of  cast  iron  and  weighs 
136  Ib.    What  will  be  the  weight  of  its  pattern  if  it  is  con- 
structed of  white  pine  ? 


PATTERN  MAKING  AND   FOUNDRY  WORK       89 


TT 


7.  The  pattern  for  a  hand  wheel  12^  in.  in  diameter  is 
made  of  white  pine  and  weighs  \  Ib.   What  will  the  casting 
weigh  if  it  is  made  of  cast  iron  ?    (See  table  of  weights.) 

8.  How  much  cast  iron  must  be  melted  for  10  pulleys 
with  8f  in.  face  and  20|  in.  diameter,  if  the  pattern  is 
made  of  cherry  and  weighs  8  Ib.? 

9.  The  side  spindle  pulley  for  a  planer  and  matcher  is 
cast  iron  ;  the  weight  of  its  mahogany  pattern  is  2£  Ib. 
How  much  metal  must  be  melted  for  36  pulleys,  if  If  Ib. 

is  deducted  from  each  for  the    , — , 

hole  cored  out  for  the  shaft  ? 

10.  A  lever-arm  pattern  made 
of  white  pine  weighs  1^  Ib.    In 
the  larger  end  there  is  an  open- 
ing 3^  in.  in  diameter  that  re- 
quires a  core  5£  in.  long.     If 
we  make  12  iron  castings,  how 
much  iron  must  we  have  ? 

11.  A  pattern  for  a  rectan- 
gular iron  casting  is  3  by  3  by 

10  in.    What  is  the  weight  of  the  casting,  cast  iron  weigh- 
ing 26  Ib.  per  cubic  inch? 

12.  The  pattern  for  a  hollow  brass  cylinder  is  10  in.  long, 
2^  in.  in  diameter,  and  has  core  prints  1^  in.  in  diameter 
on  each  end.    If  brass  weighs   488.75  Ib.  per  cubic  foot, 
what  is  the  weight  of  the  casting  ? 

13.  What  is  the  weight  of  an  engine-crank  disk  (cast 
iron),  if  its  pattern  is  4  in.  thick  and  30  in.  in  diameter, 
20  Ib.  being  deducted  for  openings  ? 

14.  The  pattern  for  a  brass  bushing  is  made  of  mahogany 
and  weighs  1£  Ib.    If  the  specific  gravity  of  mahogany  is 
.560  and  that  of  brass  7.82,  what  is  the  weight  of  the  cast- 
ing, the  core  equaling  ^  of  the  volume  of  the  pattern  ? 


FIG.  48.    SURFACE  PLATE 


90  SHOP  PROBLEMS  IN  MATHEMATICS 

15.  At  3  cents  per  lb.,  what  will  be  the  cost  of  25  cast- 
iron  surface  plates  16  in.  by  18  in.  by  f  in.  thick,  ribbed 
f  in.  deep,  as  shown  in  Fig.  48  ? 

16.  The  pattern  for  a  brass  bushing  weighs  3  lb.,  and 
the  core  prints  are  3J  in.  in  diameter  and  10  in.  long. 

What  is  the  weight  of  the 
bushing  if  the  pattern  is  made 
of  cherry,  the  specific  gravity 
of  which  is  .715  ? 

17.  The  entire  set  of  pat- 
terns for  a  gas  engine  weighs 
3^  lb. ;  the  dry  sand  cores 

F,a.  49.    HOLLOW  CHOCK          Wei§h  ^  lb'    H°W  mUCh  WlU 

one   set   of   castings  weigh  ? 

(See  table  for  weight  of  cores.) 

18.  The  pattern  for  a  hollow  chuck  (Fig.  49)  on  the 
lathe  is  2g-  in.  in  diameter  and  2J  in.  long.   If  the  diameter 
for  the  core  prints  for  the  opening  is  If  in.,  how  many 
pounds  of  cast  iron  will  be  required  for  12  chucks  ? 

19.  The  pine  pattern  of  a  small  surface  plate  weighs 
J   lb.     How   much  would   the  I 

casting  weigh  ?    (See  table  of  « j,|o- 

weights.) 

20.  If  the  pattern  for  the  face 
plate  shown  in  Fig.  50  is  made  | 

of  cherry  and  weighs  J  lb.,  how  [^ Y/Q  ,| 

much  cast  iron  will  be  required 

0     Jr.  FIG.  50.    FACE  PLATE 

to-  make  36  castings  ?    (Cherry 

weighs  38  lb.,  cast  iron  450  lb.  per  cubic  foot.) 

21.  If  brass  weighs  .31  lb.  per  cubic  inch  and  mahogany 
weighs  51  lb.  per  cubic  foot,  how  many  pounds  of  brass 
will  be  required  for  a  solid  brass  roller  if  the  mahogany 
pattern  weighs  4^  lb.  ? 


2->'^ 


PATTERN   MAKING  AND   FOUNDRY  WORK        91 


22.  If  white  pine  weighs  34  Ib.  per  cubic  foot  and  cast 
iron  weighs  450  Ib.,  how  much  iron  must  be   melted  to 
make  a  solid  casting  from  a  pattern  that  weighs  8  Ib.? 
from  a  pattern  that  weighs 

12  Ib.  ?  from  a  pattern  that 
weighs  15  Ib.  ?  (No  allow- 
ance for  waste.) 

23.  The  pattern  for  a  cast- 
iron  flanged  pulley  weighs 
1|  Ib.  If  the  pattern  is  made 
of  cherry  and  there    is   an 
opening  through  the  casting 
1^  in.  in  diameter  and  6^  in. 
long,  how  much  metal  is  re- 
quired ?  (Disregard  the  core 
print.) 


FIG.  51.    CONE  PULLEY 

24.  How  much  cast  iron  must  be  melted  for  50  cone 
pulleys  like  those  in  Fig.  51? 

PRESSURE  IN  MOLDS 

74.  Bursting  of  molds.    It  is  important  that  the  student 
should  be  able  to  figure  out  the  pressure  exerted  upon  the 
mold  by  different  metals,  in  order  to  take  proper  care  in 
preventing  the  molds  from  bursting  or  leaking. 

75.  Laws  of  pressure.    The  laws  of  pressure  are : 

(1)  Fluid  pressure  at  a  point  in  a  liquid  at  rest  is  the 
same  in  all  directions. 

(2)  The  total  pressure  exerted  on  any  surface  is  equal  to 
the  product  of  the  area  of  that  surface,  the  weight  of  a  cubic 
inch  of  that  liquid,  and  the  height  of  the  liquid  above  the 
center  of  the  surface.    If  the  given  surface  is  irregular,  the 
center  of  gravity  must  be  taken  instead  of  the  center  of 
surface. 


92 


SHOP  PROBLEMS  IN  MATHEMATICS 


EXAMPLE.  Denoting  the  pressure  by  P,  area  by  A,  weight 
per  cubic  inch  by  w,  the  height  of  the  column  of  liquid  by 
A,  find  the  pressure  exerted  on  the  surface  db  (see  Fig.  52) 
by  the  melted  iron  when  h  =  5",  w  =  .26  lb.,  and 
A  =  13"  x  10"  =  130  sq.  in. 
P  =  130  x  .26  x  5  =  171  lb'. 

Therefore  a  force  of  171  lb.  tends  to  lift  the  cope  from 
the  drag. 

Find  the  pressure  exerted  sidewise  on  ac.    The  center  of 
gravity  of  this  surface  is  at  G,  11£"  below  the  top.    Strip 
ac  is  10"  wide  and  13"  deep ;  then 

P  =  130  x  .26  x  11 J  =  388.70  lb. ; 

i.e.  388.7  lb.  is  pressing 
toward  the  left  and  the 
right,  tending  to  force  the 
flask  apart  sidewise.  Simi- 
larly, the  pressure  on  cd 
may  be  found  by  taking 


FIG.  52.    ANGLE  IRON  IN  MOLD 


h  =  18"  and  e  =  1". 

76.  Sprue.  The  channel 
through  which  the  metal  is 
let  into  the  mold  is  called 

a  sprue.    The  top  of  the  sprue  is  sometimes  spoken  of  as 

a  pouring  basin.    (See  Fig.  52.) 

77.  Pressure   head.    The  distance  from  the  top  of  the 
metal  in  the  mold  to  the  level  of  the  sprue  is  called  the 
pressure  head. 

To  insure  sound  castings  the  pressure  head  should  be  sufficiently 
great  to  drive  out  all  gases  produced  by  the  hot  metal.  These  gases 
escape  through  the  sand.  (See  Fig.  52.) 

78.  Cope  and  drag.    The  flask  in  which  a  mold  is  made  is 
generally  in  two  parts.    The  upper  part  is  called  the  cope. 
The  lower  part  is  called  the  drag. 


PATTERN  MAKING  AND  FOUNDRY  WORK       93 


PROBLEMS 

1.  Denoting  the  pressure  in  the  mold  by  P,  the  area  by 
A,  the  weight  per  cubic  inch  by  w,  and  the  height  of  the 
column  of  the  liquid  by  h,  write 

the  formula  for  P  in  terms  of 
A,  w,  and  h. 

2.  The  top  of  a  surface  plate 
measures  16"  x  18".    What  is 
the   upward    pressure   exerted 
on  the  cope  if  the  cope  is  8" 
deep  ?    (See  Fig.  48.) 

3.  A  large  face  plate  18"  in 
diameter  has  a  rim  1J"  wide. 
The  distance  from  the  top  of 

the  mold  to  the  top  of  the  face  plate  is  9".  Find  the  up- 
ward pressure.  What  is  the  total  lateral  (side)  pressure 
exerted  by  the  rim  ? 

4.  A  core  is  made  as  shown  in  Fig.  53.    The  distance  of 
the  surface  a  below  the  top  of  the  liquid  iron  is  6",  and  of 
the  surface  b,  6f ".    Is  the  balance  of  pressure  upward  on  b 

or  downward  on  a  ? 


FIG.  53.    SHOWING  CORE 
IN  MOLD 


J 


5.  A  block  of  the  size  shown 
in  Fig.  54  is  to  be  cast.   If  the 
depth  of  the  drag  is  7",  which 
face  should  be  up  to  give  the 
least  lifting  pressure  ? 

6.  A  water  pipe  3  ft.  in  out- 
side diameter,  10  ft.  long,  and 

1"  thick  is  cast  on  end.  The  top  of  the  pipe  is  1  ft.  below 
the  top  of  the  pouring  basin.  Find  the  total  pressure  exerted 
on  the  bottom  of  the  mold  and  the  total  side  pressure  tend- 
ing to  burst  the  mold. 


FIG.  54.    BLOCK 


94  SHOP  PROBLEMS  IN  MATHEMATICS 

7.  Given  a  crucible  measuring  10"  in  diameter  at  the 
top,  11"  at  the  middle,  9"  at  the  bottom,  and  14"  high  on 
the  inside.   Approximately,  how  many  pounds  of  brass  will 
it  hold,  brass  weighing  .29  Ib.  per  cubic  inch.    How  many 
pounds  of  iron,  weight  per  cubic  inch  .260  ?    How  many 
pounds  of  lead,  weight  per  cubic  inch  .410? 

8.  A  graphite  crucible  is  4^"  inside  diameter  at  the  top, 
3"  at  the  bottom,  and  7^"  high.    What  approximate  weight 
of  brass  will  it  hold  ? 

THE  CUPOLA  AND  BLAST  FURNACE 

79.  Total  charge.    The  total  weight  of  iron  put  into  the 
cupola  during  the  heat  is  called  the  "  total  charge." 

80.  Overcharge.    The  weight  of  iron  in  excess  of  what 
is  required  to  pour  the  molds  is  called  the  "  overcharge." 

81.  Shrinkage.    The  total  loss  of  weight  of  iron  by  burn- 
ing and  otherwise  is  the  shrinkage.    The  shrinkage  is  found 
by  deducting  from  the  total  charge  the  weight  of  the  cast- 
ings delivered,  the  sprues,  poor  castings,  and  the  excess  of 
iron  after  all  the  molds  have  been  poured. 

The  ratio  of  fuel  to  metal  is  valuable  in  figuring  out 
costs  and  in  finding  the  quantities  of  fuel  for  subsequent 
charges.  The  ratio  varies  greatly  with  different  cupolas 
and  with  different  kinds  of  work.. 

PROBLEMS 

1.  A  cupola  measures  24"  inside  diameter  and  8"  from 
the   sand  bottom  to  the  bottom  of  tuyeres.    How  many 
pounds  of  iron  will  it  hold  ? 

For  parts  of  cupola,  see  Fig.  55. 

2.  How  many  times  would  a  cupola  like  the  preceding  have 
to  be  tapped  to  draw  off  4800  Ib.  of  iron  ?    Allow  a  15  per 
cent  margin  in  the  bottom  of  the  cupola  for  slag  and  safety. 


PATTERN   MAKING  AND   FOUNDRY  WORK        95 

3.  What  diameter  and  height  should  a  ladle  be  to  hold 
the  metal  from  one  of  these  taps  ?  The  ratio  of  the  height 
to  the  diameter  of  the  ladle  may  be  1^  to  1. 


FIG.  55.    CUPOLA 

4.  A  cupola  melts  43,000  Ib.  of  iron  with  5400  Ib.  of 
fuel.    What  is  the  ratio  of  fuel  to  iron  ? 

5.  A  cupola  measures  34"  inside  diameter  and  14"  from 
bottom  plate  to  bottom  of  tuyeres.    Allow  4"  for  the  bed 


96  SHOP  PROBLEMS  IN  MATHEMATICS 

lining.    How  many  pounds  of  melted  iron  will  this  cupola 
hold  before  it  is  necessary  to  tap  ? 

6.  A  cupola  melts  3650  Ib.  of  iron  with  550  Ib.  of  fuel. 
What  is  the  ratio  of  fuel  to  iron  ? 

7.  It  is  desired  to  melt  a  charge  of  2400  Ib.  of  iron. 
How  much  fuel  would  be  required  at  the'  ratio  found  in 
Problem  6  ? 

8.  Allowing  $20  a  ton  for  pig  iron  and  $4.75  a  ton  for 
coke,  what  value  is  represented  in  the  materials  mentioned 
in  each  of  the  two  preceding  problems  ? 

9.  A  cupola  melts  8800  Ib.  of  iron  with  850  Ib.  of  coke. 
What  is  the  ratio  of  fuel  to  iron  ? 

10.  The  first  charge  of  fuel  in  a  cupola  is  500  Ib.  and 
the  first  charge  of  iron  is  1500  Ib.    What  is  the  ratio  ? 

1 1.  The  second  charge  of  fuel  is  60  Ib.  and  of  iron  1200  Ib. 
What  is  the  ratio  ? 

The  ratio  of  total  charges  is  shown  in  Problem  9. 

12.  A  cupola  melts  6800  Ib.  of  iron  with  720  Ib.  of  coke. 
What  is  the  ratio  of  fuel  to  iron  ? 

13.  A  cupola  melts  12,000  Ib.  of  iron  with  2200  Ib.  of 
coke.    What  is  the  ratio  of  fuel  to  iron  ? 

14.  A  cupola  melts  37,700  Ib.  of  iron  with  3550  Ib.  of 
coke.    What  is  the  ratio  of  fuel  to  iron  ? 

Problems  15,  16,  17,  and  18  are  based  on  the  same  cupola. 

15.  A  cupola  measures  25"  inside  diameter  and  8"(average) 
from  sand  bottom  to  bottom  of  tuyeres.    How  much  melted 
iron  can  it  hold  before  the  iron  runs  out  through  the  tuyeres  ? 

16.  The  thickness  of  the  fire-brick  lining  is  8",  inside 
diameter  25",  and  height  of  fire-brick  lining  15  ft.    Deduct 
4   sq.  ft.  for  charging  door.    Find  the  number  of  brick 
9"  x  4"  x  2"  necessary  to  line  the  cupola. 


PATTERN  MAKING  AND  FOUNDRY  WORK       97 

17.  The  first  charge  of  fuel   was  400  Ib.  and  of  iron 
1600  Ib. ;   the  second  charge   of  fuel  100  Ib.  and  of  iron 
900  Ib.    What  was  the  ratio  of  fuel  to  iron  in  the  first  and 
second  charges  ? 

18.  The  total  iron  melted  was  3400  Ib.  and  the  fuel  used 
was  610  Ib.    What  was  the  total  ratio  of  fuel  to  iron  ? 

19.  The  first  charge  of  fuel  in  a  Colliau  cupola  is  250  Ib. 
and  the  first  charge  of  iron  1000  Ib.    What  is  the  ratio  of 
fuel  to  iron  ? 

20.  The  second  charge  of  fuel  is  70  Ib.  and  of  iron  700  Ib. 
What  is  the  ratio  ? 

21.  The  total  charge  of  fuel  is  390  Ib.  and  of  iron  2400  Ib. 
What  is  the  final  ratio  of  fuel  to  iron  ? 

22.  A  charge  is  to  be  made  up  of  different  kinds  of  iron  as 
follows  :  Emporium,  25  per  cent;  Saxon,  20  per  cent ;  scrap, 
55  per  cent.  Find  weights  of  each  in  a  total  charge  of  3600  Ib. 

23.  A  total  charge  of  iron  is  composed  of  8500  Ib.  of 
Genesee  pig  iron,  8500  Ib.  of  Susquehanna,  and  9400  Ib.  of 
scrap.   WThat  is  the  percentage  of  each  ?    If  the  1st,  2d,  3d, 
and  4th  charges  of  Genesee  are  respectively  2000,  1050, 
1100,  1000  Ib.,  what  will  the  weights  of  scrap  be  if  added 
as  in  the  above  ? 

24.  We  desire  to  make  the  following  mixture  for  our 
foundry  castings  :   Tonawanda  No.  1,  30  per  cent ;   Erie 
No.  2,  25  per  cent  ;  and  scrap,  45  per  cent.    What  weights 
will  be  needed  of  each  for  a  heat  of  38,700  Ib.  ? 

25.  A  cupola  charge  consists  of  26,600  Ib.  of  iron  costing 
$16  a  ton  and  4500  Ib.  of  coke  costing  $4.25  a  ton.    What 
is  the  cost  per  pound  of  iron  ? 

26.  In  Problem  25,  if  it  requires  a  melter  at  $3  a  day 
and  two  helpers,  each  at  $2  a  day,  what  will  be  the  total 
cost  per  pound  for  material  and  labor  ? 


98  SHOP  PROBLEMS  IN  MATHEMATICS 

27.  A  molder  with  one  helper 'makes  one  mold  a  day. 
If  the  casting  weighs  500  Ib.  and  the  wages  of  both  molder 
and  helper  are  $5.25  a  day,  what  must  be  charged  in  order 
to  give  1  cent  per  pound  profit  ? 

28.  A  molder,  without  a  helper,  earning  $4  per  day, 
makes  20  molds  in  a  day  ;    every  mold  yields  11  Ib.  of 
good  castings.    If  the  cost  of  material  is  the  same  as  that 
found  from  Problem  26,  and  the  manufacturer  wishes  to 
make  1  cent  a  pound  profit,  what  must  he  charge  per  pound 
for  these  castings  ? 

29.  A  blast  furnace  requires   2420  Ib.  of  hard  coal  to 
produce  1  T.  of  iron.    The    ore  used  contains   59.54  per 
cent  iron.    What  proportion  of  coal  to  iron  ore  must  be 
charged  into  the  furnace  ?    (Use  long  ton  of  2240  Ib.) 

30.  A  blast  furnace  produces  1357  tons  of  iron  every 
week.    How  much  coal  and  iron  ore  would  it  consume  ? 

31.  A  blast-furnace  charge  may  be  4000  Ib.  of  ore  con- 
taining 60.7  per  cent  of  iron,  900   Ib.  of   limestone,  and 
2300  Ib.  of  coke.    What  is  the  percentage  of  each  per  ton 
of  pig  iron  produced?    (Long  ton.) 

32.  At  $1.50  a  ton  for  coke,  what  would  be  the  cost  of 
fuel  sufficient  to  melt  a  day's  run  of  200  T.  of  iron,  when 
made  from  the  above  ore  ?    (Long  ton.) 

33.  How  much  iron  ore,  limestone,  and  coke  would  be 
needed  to  produce  200  T.  of  iron  from  the  above  mixture  ? 
(Long  ton.) 

COST  OF  MATERIAL  AND  LABOR 

82.  Estimating.  In  estimating  the  time  required  to  do 
work  or  "  jobs  "  in  manufacturing  plants,  experience  shows 
the  estimator  about  how  much  time  should  be  allowed  for 
every*operation  performed  in  his  shop.  To  ascertain  the 


PATTERN  MAKING  AND  FOUNDRY  WORK        99 

cost  of  a  piece  of  work,  add  to  the  cost  of  material  the 
cost  for  labor  in  every  operation,  the  fixed  or  office  charge, 
and  the  profit. 

In  machine  work,  for  example,  on  any  one  job  there  will 
be  lathe  work  (turning  and  thread  cutting),  milling,  drill- 
ing, etc.,  the  time  for  all  of  which  must  be  estimated  if  a 
price  is  to  be  named  before  the  job  is  completed. 

PROBLEMS 

1.  Soft  solder  is  composed  of  2  parts  tin  and  1  part  lead. 
How  many  pounds  of  each  are  required  to  make  25  Ib. 
of  solder  ? 

2.  If  a  solder  is  composed  of  H  parts  tin  and  1  part  lead, 
how  many  pounds  of  each  are  required  for  25  Ib.  of  solder  ? 

3.  At  35  cents  a  pound  for  tin  and  4  cents  a  pound  for 
lead,  which  of  the  compositions  in  Problems  1  and  2  is 
cheaper  ? 

4.  Silver  solder  is  1  part  copper  to  2  parts  silver.    How 
much  of  each  is  in  6  oz.  of  solder  ? 

5.  If  silver  solder  had  the  mixture  of  1  part  brass,  1  part 
copper,  and  19  parts  silver,  how  much  of  each  would  there 
be  in  6  oz.  of  solder  ? 

6.  At  15  cents  a  pound  for  brass,  17  cents  for  copper, 
and  55  cents  an  ounce  for  silver,  which  of  the  compositions 
given  in  Problems  4  and  5  is  cheaper  ? 

7.  From  a  solid  bar  1^"  in  diameter  a  bolt  is  turned  with 
the  following  dimensions:  f"  in  diameter;  4"  long  under 
the  head;  head  f"  thick  and  lTy  square.  What  proportion 
of  stock  is  wasted  ? 

8.  What  will  be  the  cost  of  a  solid  brass  sphere  5  in.  in 
diameter,  if  copper  costs  20  cents  per  pound  and  zinc  6£ 
cents  ?    The  weight  of  brass  is  488.75  Ib.  per  cubic  foot 
and  the  proportion  is  2  of  copper  to  1  of  zinc. 


100  SHOP  PROBLEMS  IN  MATHEMATICS 

9.  Brass  contains  ,65  per  cent  copper  and  35  per  cent 
zinc.    If  copper  is  selling  at  17  cents  a  pound  and  zinc 
at  5£  cents,  for  what  ought  brass  to  sell  when  2  cents  a 
pound  is  allowed  for  casting  the  brass  ? 

10.  If  bronze  contains  85  per  cent  copper,  9  per  cent 
zinc,  and  6  per  cent  tin,  what  is  it  worth  per  pound  when 
copper  sells  for  18  cents,  zinc  for  6  cents,  and  tin  for  35 
cents  ? 

11.  A  machinist  receiving  $3.25  per  day  is  given  a  10 
per  cent  increase.   A  few  months  later  he  is  cut  10  per  cent. 
What  is  his  wage  now  ?    Why  is  it  not  the  same  as  before  ? 

12.  A  shop  has  been  paying  its  men  for  ten  hours'  work, 
but  it  is  now  obliged  to  pay  them  the  same  day's  wage 
for  an  eight-hour  day.    What  per  cent  increase  does  each 
man  receive  ? 

13.  What  would  men  receiving  the  following  per  hour  in 
a  ten-hour  day  get  per  hour  in  an  eight-hour  day,  the  total 
day's  wage  to  be  the  same  :  20  cents,  22 J,  25,  27 J,  30,  33, 
35,  37£  ? 

14.  Make  a  table  of  wages  based  on  37^  cents  per  hour 
for  every  successive  20  inin.  of  an  eight-hour  day,  —  that  is, 
20  min.,  40  min.,  1  hr.,  1  hr.  and  20  min.,  etc. 

15.  If  the  time  cards  for  a  piece  of  work  show  3  hr.  and 
15  min.  lathe  work,  12  hr.  and  15  min.  planing,  and  45  min. 
vise  work,  what  is  the  total  cost  of  the  job  ?   Allow  $2.75 
for  lathe  work,  $2.45  for  planing,  and  $2.30  for  vise  work 
per  day  of  8  hr. 

16.  If  a  man  makes  300  bolts  at  3  cents  each,  175  at  2J 
cents,  and  560  at  3^  cents,  what  are  his  total  earnings  ? 

17.  If  a  man  works  5  da.,  9  hr.  a  day  and  5f  hr.  over, 
what  will  his  week's  wages  be  at  33^  cents  an  hour  ? 

18.  What  wages  per  day  of  9  hr.  would  be  equivalent  to 
$27.25  per  week  ? 


PATTERN  MAKING  AND  FOUNDRY  WORK  101 

19.  If  a  man  works  65^  hr.  in  one  week,  54  of  which  is 
regular  day  work  at  29|  cents  an  hour,  and  the  remainder 
is   overtime,  counted  as  time  and  one  half,  what  are  his 
week's  wages  ? 

20.  An  apprentice  receives  $6.50  a  week.    In  that  time 
he  makes  55  large  cap  screws.    A  skilled  mechanic  at  30 
cents  an  hour  (nine-hour  day)  can  make  25  of  the  same 
pieces  a  day.    Whose  labor  is  cheaper  ? 

21.  A  screw  made  of  brass  has  a  head  T7ff"  in  diameter 
by  J"  thick,  and  a  body  T3ff"  in  diameter  by  T\"  long.   What 
will  be  its  approximate  weight  ?   The  weight  of  a  cubic  inch 
of  brass  is  0.31  Ib. 

22.  What  per  cent  of  the  material  for  the  screw  described 
in  Problem  21  is  wasted  if  the  same  is  made  from  a  ^"  round 
bar  of  brass,  allowing  1"  in  length  for  cutting  off? 

This  does  not  allow  for  thread  cut  on  screw. 

23.  If  brass  is  worth  19  cents  a  pound  and  the  scrap  is 
worth  6  cents  a  pound,  what  will  the  material  cost  for  1000 
screws  ? 

24.  If  875  screws  are  made  in  a  day  on  an  automatic 
screw  machine  at  a  labor  and  machine  cost  of  87  cents, 
what  does  each  screw  cost  ? 

Count  in  cost  of  material  as  found  above. 

25.  The  speed  lathe  has  four  speeds,  1200,  750,  400,  and 
250  R.  P.  M.    Find  the  step  which  will  give  to  a  £"  brass 
rod  a  surface  speed  of  90  ft.  a  minute  (approximately). 
Solve  for  a  £",  1",  1|",  and  2"  rod. 


OF  TH£ 

UNIVERSITY 

Of 


CHAPTER  VI 

LENGTH  OF  STOCK  FOR  FORCINGS,  STRENGTH  OF 
FORCINGS,  ALLOWANCE  FOR  SHRINK  FITS 

83.  Length  of   stock.    In  making  calculations  for  the 
amount  of  stock  required  for  a  given  forging,  the  weight 


of  the  forging  must  be  found,  and  then  an  equivalent 
weight  of  the  size  of  bar  stock  to  be  used  must  be  ascer- 
tained. In  some  cases  a  small  additional  amount  of  mate- 
rial must  be  allowed,  as,  for  example,  in  making  a  weld, 
an  allowance  in  length  equal  to  the  thickness  of  stock  is 
generally  necessary. 

102 


LENGTH  OF  STOCK  FOR  FORCINGS 


103 


FIG.  56.    ANGLE  IRON 


84.  Use  of  center  lines.  In  calculating  weights,  all  dimen- 
sions should  be  taken  from  the  axes  or  center  lines  of  the 
forging  and  not  from  the  outside  or  inside  meas- 
urements. For  example,  in  finding  the  length  of 
stock  to  make  a  right  angled  bend,  as  in  Fig.  56, 
the  length  of  the  center  line  is  4^"  +  3f "  =  8". 
The  student  should  try  to  divide  the  forging  into 
sections  of  straight,  curved, 

or  circular  parts   for  ease  in    _J — . ^ 

calculation.  -r — "^ 

A  convenient  rule  to  remem-  L 4  • 

ber  in  finding  lengths  of  circu- 
lar forgings  is  the  following: 

To  the  inside  diameter  add  one  thickness  (of  stock)  and 
multiply  by  3 ;  or,  if  greater  accuracy  is  required,  multiply 
by  2T2  (TT). 

PROBLEMS 

1.  Find  the  length  of   stock  for  a  flat  ring  J"  thick, 
f "  wide,  and  3^"  in  inside  diameter. 

2.  A  |"  bolt  is  to  be  forged  6"  long  under  the  head.    How 

much  extra  stock  must  be  allowed  for 
the  head  ? 

The  size  of  a  bolt  head  (in  the  rough)  is  equal 
to  \\  x  diameter  of  bolt  +  J".  This  applies  to 
both  square  and  hexagonal  heads.  A  finished 
head  measures  the  same.  The  thickness  of  the 
head  is  equal  to  the  diameter  of  the  bolt. 

3.  What  allowance  must  be  made  for 
the  hexagonal  head  of  a  f "  bolt  ? 

4.  A  swivel  (see  Fig.  57)  is  to  be  forged 
from  y '  square  stock.    How  much  stock 
is  required  ? 

5.  A  piston  ring  having  a  diameter  of  22"  is  -J-"  thick  and 
J"  wide.  What  will  it  weigh  at  .26  Ib.  per  cubic  inch  ? 


FIG.  57.    CHAIN 
SWIVEL  ' 


104 


SHOP  PROBLEMS  IN.  MATHEMATICS 


6.  A  crank  (see  Fig.  58)  must  be  forged  T\"  larger  all  over 
than  the  dimensions  given.    What  will  the  crank  weigh  in 
the  rough  ? 

7.  If  the  crank  is  forged  from  a  bar  2|"  square,  how  long 
a  piece  of  stock  will  be  required  ? 


3 

^ 

'     I 

1  1 

Hi 

FIG.  58.    DOUBLE-THROW  CRANK 

8.  If  these  cranks,  when  made  in  lots  of  15,  are  forged  at 
cents  per  pound,  what  would  one  crank  cost  ? 

9.  Find  the  length  of  1^"  square  stock  necessary  to  forge 
the  crank  shown  in  Fig.  59,  all  the  dimensions  given  being 
finished  measurements.    Allow  ^"  for  finish. 

10.  The  leg  of  an.  andiron  (see  Fig.  60)  is  made  from 
1J"  square  iron.    Find  the  length  of  stock  required. 

r2: 


-  — 

sW 

-     /  / 

i 

—f-     MOO 

-1-  

—  *  

tid 

u-  je: 

FIG.  59.    SINGLE-THROW  CRANK 

11.  A  disk  3^"  in  diameter  and  1"  thick  is  required  for 
the  machine  shop.    The  largest  round  bar  in  the  shop  is  2£" 
in  diameter.    How  long  a  piece  of  this  bar  must  be  cut  off 
to  make  the  disk  ? 

12.  A  ring  10"  in  outside  diameter,  -J"  thick,  and  3"  wide 
is  required.   The  nearest  size  of  flat  bar  in  stock  is  2|-"  x  !£"• 
How  long  a  piece  of  the  bar  must  be  used  ? 


LENGTH  OF  STOCK  FOR  FORCINGS 


105 


13.  An  eyebolt,  shown  in  Fig.  61,  must  be  forged  from 
1"  X  1J"  stock.    How  long  a  piece  will  be  needed  ? 

14.  A  flat  ring  has    the   following    dimensions  :    inside 
diameter  6",  outside  diameter  8"   thickness     ".    Select  a 


suitable  size  of  stock,  not  over  4"  wide,  and  ascertain  the 
amount  required. 

Add  about  \  the  width  of  stock  to  the  length  calculated,  to  allow 
for  welding. 

15.  In  every  2000  Ib.  of  coal  there  are  117.5  Ib.  of  ash. 
What  is  the  per  cent  of  ash  ? 

16.  A  welded  iron  stay  may  be  counted  75  per  cent  as 
strong  as  the  original  bar.    Allowing  the  tensile  strength  of 
a  1"  round  bar  to  be  35,000  Ib.  per  square  inch,  what  will  be 
its  strength  after  it  has  been  welded  ? 

17.  What  must  be  the  size  of  a  welded  bar,  to  be  as 
strong   as   a   1"  bar    before 

welding?    A  bar  is   1£"  in    KH  ''H 7 


diameter  at  the  weld.  How 
small  may  it  be  outside  the 
weld  to  be  of  the  same 
strength  ? 


FIG.  61.    EYEBOLT 


18.  An  axle  is  7  ft.  long  and  averages  5"  in  diameter. 
Compute  its  weight,  assuming  steel  to  weigh  489  Ib.  per 
cubic  foot. 


FIG.  62.    GROUP  OF  FORCINGS 
106 


LENGTH  OF  STOCK  FOR  FORCINGS  107 

19.  Find  the  length  of  stock  necessary  to  make  each  of 
the  forgings  shown  in  Fig.  62. 

Select  sizes  of  stock  which  most  nearly  conform  with  the  largest 
part  of  the  forging. 

20.  A  T  rail  (see  Fig.  63)  30  ft.  long  weighs  80  Ib.  to 
the  yard.    How  many  rails  in  a  ton  of  2240  Ib.  ?    What 
would  each  rail  cost  at  $27.60  per  ton  (2240  Ib.)? 

21.  What  would  the  T  rails  cost  for  a  section  of  railroad 
2  mi.  long  ? 

22.  If  a  70-lb.  rail  instead  of  an 
80-lb.  were  used,  how  much  would 
be  saved  in  a  1000-mi.  railroad  ? 

23.  If  a  girder  rail  (see  Fig.  63) 
weighs  104  Ib.  to  the  yard  and  is 
60  ft.  long,  what  is  its  entire  weight  ? 

What  does  it  cost  at  $34.70  a  ton     Flu'  c3'   T  AND  GlRDER 

RAILS 
(2240  Ib.)  ? 

24.  What  would  be  the  cost  of  10  mi.  of  railroad  made 
from  the  girder  rail  ? 

25.  How  much  more  expensive  is  the  1000-mi.  railroad 
made  with  the  girder  rail  than  if  made  with  the  T  rail  ? 

26.  The  long  arm  of  a  shear  is  26"  and  the  short  arm  9". 
If  a  force  of  1000  Ib.  is  applied  at  the  end  of  the  long  arm, 
what  sized  square  bar  would  the  shear  cut  when  placed  6" 
from  the  pivot  ?    What  width  of  a  f "  flat  bar  would  the 
shear  cut  ? 

27.  A  punch  has  a  leverage  of  2f "  to  16".    Allowing  the 
shearing  force  of  iron  to  be  50,000  Ib.  per  square  inch,  how 
much  force  applied  at  the  end  of  the  long  arm  is  required 
to  punch  a  -J-"  hole  ?    How  much  work  is  done  if  the  plate 
is  y  thick  ? 

Work  done  is  equal  to  the  force  multiplied  by  the  number  of  feet 
through  which  the  force  acts. 


108  SHOP  PROBLEMS  IN  MATHEMATICS 

28.  A  blacksmith's  shear  and  punch  has  a  leverage  on  the 
cutting  arm  of  10  to  1.    What  force  at  the  power  end  would 
be  necessary  to  shear  a  bar  of  iron  T»ff"  x  l£"?    Shearing 
stress  for  iron  is  50,000  Ib.  per  square  inch.    How  much 
force  for  stock  J"  x  |",  ft"  X  |",  f"  X  Ifc",  i"  X  1"?  What 
is  the  work  done  in  each  case  ? 

29.  Soft  steel  contains  approximately  0.37  per  cent  of 
silicon.    In  1  T.  of  steel  how  much  is  silicon  ? 

30.  One  grade  of  tool  steel  contains  1.2  per  cent  carbon. 
How  many  pounds  of  carbon  are  contained  in  200  Ib.  of  this 
steel ? 

31.  A  Bessemer  converter  contains  10  T.  of  liquid  iron 
to  which  it  is  desired  to  add  0.5  per  cent  manganese.    How 
much  ferromanganese  containing  20  per  cent  of  manganese 
must  be  added  ? 

32.  A  chrome-steel  armor  plate  containing  1.75  per  cent 
of  chromium  and  weighing  8  T.  is  to  be  cast.    How  much 
ferrochrome  containing  65  per  cent  of  chromium  must  be 
used? 

33.  Nickel   steel   for   automobile   construction    contains 
3.40  per  cent  of  nickel.    If  a  ladle  holds  3700  Ib.  of  steel, 
how  many  pounds  of  nickel  must  be  added  ? 

34.  Nickel  costs  30  cents  a  pound.  How  much  more  would 
the  materials  cost  in  37,000  Ib.  of  nickel  steel  than  in  the 
same  amount  of  mild  steel  at  $20  a  ton  (2240  Ib.). 

Nickel  steel  is  3.5  per  cent  nickel. 

35.  A  nickel-chrome  steel  contains  .45  per  cent  of  chro- 
mium and  1.25  per  cent  of  nickel.    How  much  of  each 
(chromium  and  nickel)  are  contained  in  12  T.  (2000  Ib.) 
of  this  composition? 

36.  A  15-lb.  tool  steel  die  is  to  be  hardened.   When  ready 
to  be  put  into  the  water  its  temperature  is  1450°  F.   How 


LENGTH  OF  STOCK  FOR  FORCINGS  109 

large  must  the  tank  be  in  order  that  the  temperature  of  the 
water  (when  cooling  the  die)  may  not  rise  above  100°  F.  ? 
The  temperature  of  the  water  at  first  is  50°  and  the  specific 
heat  of  steel  is  .117. 

37.  Suppose  the  above  die  is  to  be  hardened  in  oil  in 
which  the  temperature   may   rise   to  300°  F.    How  many 
gallons  of  oil  would  be  required  ?    The  specific  heat  of  the 
oil  is  .310. 

38.  If  500  steel  gears  weighing  -^  Ib.  each  are  hardened 
by  cooling  from  a  temperature  of  1475°  F.  to  100°  F.  in 
one  day,  how  much  heat  will  be  carried  away  by  the  cool- 
ing substance  ?  How  many  pounds  of  coal  would  be  required 
to  provide  this  amount  of  heat  if  there  were  14,000  heat 
units  in  each  pound  ?   How  many  pounds  if  but  60  per  cent 
of  the  heat  were  available  ? 

39.  How  many  pounds  of  steel  would  a  10-gallon  pail  of 
water  harden  if  we  allow  the  water  to  rise  from  50°  to  180° 
F.  ?   The  steel  is  to  be  cooled  from  1350°  to  the  temperature 
of  the  water,  and  the  specific  heat  of  water  is  .117. 

85.  Shrink  and  force  fits  for  shafts  and  pulleys.  In  a  loose 
fit  the  diameter  of  the  shaft  may  be  from  .01"  to  .02"  smaller 
than  the  diameter  of  the  hole.  For  a  bearing  the  difference 
between  shaft  and  pulley  may  be  from  .003"  to  .006"  per 
inch  of  diameter.  For  a  sliding  fit  the  shaft  may  be  .001" 
undersize  per  inch  of  diameter.  For  an  easy  driving  fit  the 
hole  and  shaft*  may  be  as  nearly  the  same  size  as  possible. 
For  driving  fits  .001"  per  inch  of  diameter  is  a  fair  allow- 
ance. The  shaft  in  this  case  must  be  larger  than  the  hole. 
For  pressed  fits  the  allowance  is  the  same  as  for  driving 
fits.  For  shrink  fits  the  shaft  may  be  from  .001"  to  .002" 
oversize  per  inch  of  diameter. 

It  is  understood  that  the  variation  is  generally  made  in  the  shaft 
and  not  in  the  hole,  as  the  latter  is  often  determined  by  the  size  of 
the  reamer. 


110  SHOP  PROBLEMS  IN  MATHEMATICS 

The  term  "  nominal  size "  is  used  to  indicate  the  size 
given  on  the  drawing  or  some  standard  dimension,  i.e.  £•", 
i",  II",  etc. 

PROBLEMS 

1.  A  driving-wheel  tire  is  to  be  shrunk 'upon  the  wheel 
center.    The  tire  is  bored  smaller  than  the  outside  of  the 
wheel  center  by  .001"  for  every  inch  of   diameter.    How 
much  smaller  should  a  60"  tire  be  turned  if  its  rim  is  2^" 
thick  ? 

2.  How  much  shorter  would  the  inner  circumference  of 
the  tire  be  than  the  outer  circumference  of  the  wheel  center  ? 

3.  The  hub  of  a  car  wheel  is  bored  to  5.377".    How  much 
larger  should  the  axle  be  turned  for  a  pressed  fit  ? 

4.  A  shaft  is  1.4375"  in  diameter.    To  what  size  must  a 
collar  be  bored  to  give  a  solid  fit  ? 

5.  A  shaft  nominally  3.777"  slides  freely  in   a  pulley. 
What  should  be  the  exact  size  of  each  ? 

6.  Using    the    formula    .001  d  +  .0005  =  allowance    for 
press  fits,  where  d  =  diameter  in  inches,  find  the  diameters 
of  shafts  of  the  following  nominal  sizes :  f",  1",  1+",  1 J", 
li",  If",  li". 

7.  Iron  expands  0.0000066  of  its  length  for  each  degree 
of  temperature  (Fahrenheit)  increased.    If  a  steel  shaft  5  ft. 
long  at  70°  F.  is  heated,  in  turning,  to  300°  F.,  how  much 
longer  will  it  be  when  hot  ? 

8.  A  line  shaft  is  100  ft.  long  at  the  temperature  of  40°  F. 
How  long  would  it  be  at  80°  or  summer  temperature  ? 

9.  How  much  space  should  be  left  between  T  rails,  30  ft. 
long,  laid  in  the  winter  at  a  temperature  of  40°  F.  ? 


CHAPTER  VII 


SPEEDS  OF  PULLEYS,  SHAFTS,  AND  GEARS 

Before  reading  this  chapter  the  student  should  be  familiar  with 
sections  55-60  on  the  speeds  of  pulleys.  The  problems  assume  a 
knowledge  of  the  general  processes  of  machine  work,  but  a  brief 
description  of  some  terms  is  given  below. 

SPEEDS  OF  PULLEYS  AND  SHAFTS 

86.  Back  gears.   All  engine  lathes  are  provided  with  back 
gears  (see  Figs.  64  and  73),  to  increase  the  number  of  speeds 
obtainable  without  an  overin- 

crease  in  the  size  of  the  cone 
pulley. 

87.  Direct  drive.    When  the 
power  is  transmitted  directly 
from   belt   to  cone   pulley   to 
lathe  spindle,  the  term  "  di- 
rect drive  "  is  used. 

88.  Back-gear  drive.    When 
the  power  is  transmitted  from 
belt  to   cone  pulley  to   back 

gears,  thence  to  spindle,  the  term  "back-gear  drive"  is  used. 

89.  Countershaft.   The  countershaft  is  a  short  shaft  con- 
taining tight  and  loose  pulleys  and  is  placed  between  the 
main  line  shaft  and  the  machine.    Its  purpose  is  to  change 
the  speed  and  to  permit  a  convenient  location  of  machines. 
By  counterspeed  is  meant  the  speed  of  the  countershaft. 

90.  Cone  pulleys  in  geometrical  progression.    The  sizes  of 
cone  pulleys  are  so  arranged  as  to  make  the  increase  in 

111 


^E 

Back  Gear 

307: 

L 

88r. 

= 

^__ 

n 

90, 

1  — 

1H 

^z 

Lathe 

L 

32  T 

• 

_ 

Spindle 

Cone  Pulley* 

FIG.  64.    BA 

Civ 

G 

BARS 

112 


SHOP  PROBLEMS  IN  MATHEMATICS 


speeds  agree  as  closely  as  possible  with  some  geometrical 
progression.  In  the  case  of  gear  drives  for  lathes  the  speeds 
are  arranged  by  some  geometrical  ratio.  For  example,  sup- 
pose the  speeds  of  a  lathe  ran  as  follows  :  25,  35,  49,  70, 
98,  140.  By  dividing  any  one  speed  by  the  one  preceding 
we  obtain  the  geometrical  ratio  1.4.  This  ratio  is  not  exact, 
but  is  approximately  correct. 

PROBLEMS 

1.  A  motor  running  at   1800   R.  P.  M.   with   a  driving 
pulley  of  4^"  is  to  drive  a  line  shaft,  which  hangs  18"  be- 
low the  ceiling,  at  200  R  P.  M.    Can  it 
be  driven  directly  ?    Why  ? 

2.  Ascertain  all  the  pulleys  necessary 
to  drive  this  line  shaft  through  an  inter- 
mediate or  jack  shaft. 

3.  There  are  three  driven  pulleys  14" 
in  diameter  on  a  No.  1  milling-machine 
countershaft.    The  two  forward-driven 
pulleys  are  to  run  90  and  120  R  P.  M. 
and  the  reverse  105  R  P.  M.    The  line 
shaft  runs  at  200  R  P.  M.     Find  the 
diameters  of  the  line-shaft  pulleys. 

4.  The  milling-machine  countershaft 
pulleys  driven  from  a  line  shaft  are  14"  in  diameter  and  are 
to  run  100,  200,  and  150  E.  P.  M.  respectively.  Find  the  diam- 
eters of  pulleys  for  the  line  shaft  running  at  200  R P.M. 

5.  The  largest  step  on  the  cone  pulley  of  the  milling 
machine  is  10"  in  diameter  and  its  mate  on  the  counter- 
shaft is  5J".    Find  the  speed  when  the  countershaft  re- 
volves 120  times  a  minute  (see  Fig.  65). 

6.  The  cone-pulley  diameters  of  a  No.  1£  milling  machine 


Lathe  r: 
Sp.nd/et  -  - 


FIG.  65. 
CONE  PULLEYS 


are  10", 


5-J-",  and  the  corresponding  countershaft 


SPEEDS  OF  PULLEYS,  SHAFTS,  AND  GEARS    113 

diameters  are  5J-",  7",  8£",  10".  Find  the  speed  for  each 
step,  considering  the  two  forward-driven  pulleys  of  the 
countershaft  as  running  at  100  and  200  R.  P.  M.  respec- 
tively. 

There  will  be  eight  speeds. 

7.  To  what  geometrical  progression  does  this  series  of 
speeds  most  nearly  conform  ? 

8.  The  back-gear   ratio  for  milling  machine  No.  1£  is 
1  to  8.    Find  the  eight  speeds  with  back  gears  in.    Are 
they  in  the  same  geometrical  progression  as  in  Problem  7  ? 

9.  It  is  desired  to  drive  a  countershaft  of  a  12"  engine 
lathe  at  170  R.  P.  M.    The  size  of  the  driven  pulley  on  the 
counter  is  8".    The  motor  speed  is  1800  R.  P.  M.  and  its 
pulley  is  4-J-"  in  diameter.    Find  the  two  pulleys  for  the 
line  shaft.    The  pulleys  on  the  line  shaft  cannot  be  over 
37"  in  diameter. 

10.  Work  out  the  formula  for 'Problem  9,  denoting  the 
drivers  by  D  and  D1  and  the  driven  pulleys  by  d  and  d±. 

11.  A  countershaft  of  a  lathe  makes  170  R,  P.  M.    The 
diameters  of  the  steps  of  the  cones  of  both  the  lathe  and 
the  countershaft  are  as  shown  in  Fig.  65.    Ascertain  the 
spindle  speed  for  each  step. 

12.  With  the  back  gears  out,  a  lathe  spindle  makes  112 
R.  P.  M. ;  with  them  in,  it  makes  12  R.  P.  M.    What  is  the 
ratio  of  the  direct  drive  to  back-gear  drive  ? 

13.  Ascertain  the  approximate  ratio  in  the  geometrical 
progressions  of  the  following  pulley  speeds :  (a)  5,  8,  13, 
20,  33;   (b)  61,  99,  154,  242,  402. 

14.  Ascertain  the  approximate  ratio  in  the  geometrical 
progression  of  the  following  speeds,  taken  from  a  variable- 
speed  motor  drive :  54,  59,  64,  88,  108,  126,  144,  160,  172, 
192,  230,  280,  354. 


114 


SHOP  PROBLEMS  IN  MATHEMATICS 


FIG.  66 
GEARING 


15.  Considering  the  two  drives  in  Problems  13  and  14, 
which  arrangement  is  preferable,  and  why  ? 

16.  A  countershaft,  driving  a  planer,  runs  at  350  R.  P.  M. ; 
the  forward  driving  pulley  is  20"  in  diameter  and  the  back 
driving  pulley  8".    What  is  the  ratio  of  forward  to  back 
drive  ? 

GEARS 

91.  Driving  gear.    A  driving  gear  is  one  that  transmits 

the  power  to  another  gear. 

92.  Driven  gear.  A  driven  gear  is  a  gear 
to  which  the  power  is  transmitted.  If  gear 
A  (Fig.  67)  is  the  driver,  then  gear  C  is  the 
driven  gear,  and  gear  B  is  an  intermediate. 
If  the  75-tooth  gear  (Fig.  68)  is  the  driver, 
then  the  60-tooth  gear  is  also  a  driver,  and 
therefore  the  30-tooth  gear  and  the  25-tooth 

gear  are  the  driven  gears. 

93.  Intermediate  gear.   An  intermediate  gear  is  a  gear  used 
to  transmit  the  power  from  one  gear  to  another  where  these 
are  too  far  apart  to  mesh.    The 

speed  is  not  affected. 

94.  Train  of  gears.     Three  or 
more  gears  meshing  together  are 
called  a  train  of  gears.   One  or 
more  of  these  is  always  an  inter- 
mediate and  is  not  considered  in 
calculating  the  speed. 

95.  Rule  for   speed.    (1)    The 
speed  of  any  two  meshing  gears 
is  inversely  proportional   to   the 
number  of  their  teeth. 

Or  in  the  case  of  a  train  of  gears,  as  in  the  case  of  pulleys, 
the  following  rule  may  be  used : 


C 


FIG.  67.    TRAIN  OF  THREE 
GEARS 


SPEEDS  OF  PULLEYS,  SHAFTS,  AND  GEARS    115 


(2)  The  continued  product  of  the  R. 
P.  M.  of  the  first  driver  and  the  number 
of  teeth  in  every  driving  gear  is  equal 
to  the  continued  product  of  the  R.  P.  M. 
of  th  e  last  driven  gear  and  the  number 
of  teeth  in  every  driven  gear. 

PROBLEMS 


FIG.  08.    TRAIN  OF 
FOUR  GEARS 


34 


80 


FIG.  69.    GEARING  FOR 
POWER  FEED 


1.  Denoting  the  speed  of  the  driver  and  driven  gears  by 
S  and  s  respectively,  and  the  number  of  teeth  by  N  and  n, 
write  the  formula  for  s  in  terms  of  5,  TV,  and  n. 
2.  Denoting  the  R.P.M.  of  the  first  driver  by 
36  S  and  the  number  of  teeth  in  the  drivers  by  TV,  TVX, 
?     and  TV2,  the  R.  P.  M.  of  the  last  driven  gear  by  s, 
and  the  number  of  teeth  in  the  driven  gears 
by  n,  n^  and  n2,  write  the  formula  for  N  in 
terms  of  S,  TV1?  TV2,  and  s,  n,  n^  n2. 

3.  Solve  formula  in  Problem  2 
for  n  and  s. 

4.  In  Fig.  66   the  gear  of  60 
teeth  revolves   114  times    per 

minute.    Find  the  speed  of  the  driven  gear. 

5.  Gear  A  (Fig.  67)  has  69 
teeth  and  revolves  85  times  per 
minute.    Gear  C  must 
revolve    172J   times 

per  minute.    Find  the 
number  of  teeth  in  C. 

6.  The  gear  with  75 
teeth  (see  Fig.  68)  has 
a  speed  of  40  R.P.M. 
Find  the  speed  of  the 

gear  of  25  teeth.  FIG.  70.    GEARING 


116 


SHOP  PROBLEMS  IN  MATHEMATICS 


7.  Gear  A  (Fig.  69)  revolves  80  times  per  minute.    Find 
the  speed  of  shaft  F. 

8.  Select  a  gear  in  place  of  the  60-tooth  gear  that  will  cause 

F  to  turn  twice  around  while 
A  turns  around  80  times. 

9.  Select  a  "gear  in  place  of 
A  (Fig.  69)  so  that  F  will  re- 
volve Jj  as  fast  as  the  shaft 
on  which  A  is  mounted. 


Drill  Spindle 
FIG.  71.    BEVEL  AND  WORM 


n 


24 
36 
42 
46 


10.  Shaft  A  (Fig.  70)  makes  240  R.  P.  M.    Find  the  speed 
of  shaft  B. 

11.  Compute  the  ratio  of  speeds  between  shafts  A  and  B 
if  gear  A  (Fig.  70)  had  34  teeth. 

12.  The  drill  spindle  D  (Fig.  71) 
makes  110  R,.  P.  M.  Find  the  speed 
of  the  worm-wheel  shaft  E. 

13.  One    turn    of   the 
worm  wheel  E  (Fig.  71) 
raises  spindle  D  2".    Find 
the  amount  that  the  spin- 
dle  D  is   raised   by   the 
worm   wheel   in    turning 
once  around. 

Fig.  72  illustrates  the  gear- 
ing for  the  feed  of  a  22"  drill 
press.  Gear  A  on  the  drill 
spindle  drives  the  train  of 
gears  down  to  the  worm  wheel 
W.  On  the  same  shaft  with 
this  worm  wheel  is  a  spur-gear 
meshing  with  a  rack  secured 
to  the  drill  spindle,  by  which 

the  spindle  is  fed  into  the  work.    All  the  gears  in  cluster  P  are  fast  to 
the  shaft  Q,  but  those  in  cluster  0  are  arranged  in  such  a  way  that  any 


FIG.  72. 
GEARING  FOR  22"  DRILL  PRESS 


SPEEDS  OF  PULLEYS,  SHAFTS,  AND  GEARS    117 


one  may  be  made  fast  to  shaft  N.  It  then  becomes  the  driver.  Hence 
four  different  feeds  may  be  given  the  spindle  A  for  every  different 
speed  of  the  driving  pulley.  One  turn  of  worm  wheel  W  raises  the 
drill  spindle  5J". 

14.  From  the  data  given  in  Fig.  72  find  the  four  feeds  of 
the  spindle  in  decimals  of  an  inch  per  revolution. 

15.  This  same  drill  has  eight  spindle  speeds,  — 10,  16, 
26,  44,  74,  120,  200,  300.   Find  the  four  feeds  per  minute 
for  each  of  the  spindle  speeds. 


E 

1  1 
Bac 

<  G 

5sT 

c 

9 

ars 

—  1 

dtir 

1 

.m 

s 

C, 

\ 

89  T 

N 
i  — 

— 

D 

— 

897 
H 

\  n\  hp 

F 

1 

£?: 

\ 

20  T 

Spindle 

M 

58  T| 

R 

FIG.  73.    DOUBLE-BACK  GEARS 

Double-back  gearing.  A  lathe  having  a  double-back  gear  is  con- 
structed as  shown  in  Fig.  73.  Sleeve  M,  carrying  gears  F,  D,  and 
jR,  revolves  freely  on  the  lathe  spindle  L  and  is  driven  by  a  motor 
through  gear  E.  Sleeve  N  is  keyed  to  the  back-gear  shaft  S,  but 
slides  freely  from  left  to  right  and  carries  gears  E  and  C.  So  either 
gear  E  or  C  can  be  put  in  mesh  with  F  or  D  respectively,  and  the 
power  is  transmitted  from  the  motor  through  E  to  F  to  E  to  G  to  H 
and  L,  or  through  R  to  D  to  C  to  G  to  H  and  L.  With  the  back  gears 
E  and  C  both  out  of  mesh,  the  spindle  L  is  connected  with  R  by  a  pin 
or  bolt  in  the  side  of  gear  H. 


118  SHOP  PROBLEMS  IN  MATHEMATICS 

16.  If  sleeve  M  revolves  at  500  R.  P.  M.,  what  will  the 
speed  of  L  be  when  C  and  D  are  in  mesh  ?  when  E  and  F 
are  in  mesh  ? 

17.  What  is  the  ratio  of  direct  to  indirect  drive  when  C 
and  D  are  in  mesh  ?  when  E  and  F  are  in  mesh  ? 

18.  If  gear  R  has  50  teeth  and  the  gear  on  the  motor 
spindle  has  30  teeth,  what  will  be  the  speed  of  L  for  the 
following  motor  speeds  using  back  gears  E-F :  400,  500, 
600  R.P.  M.? 


CHAPTER  VIII 


CUTTING  SPEED  AND  FEED 

96.  Cutting  speed.    Cutting  speed  has  been  defined   in 
section  59,  in  connection  with  the  lathe  where  the  work 
revolves.    But  in  some  cases  the  work  does  not  revolve.   For 
example,  on  the  planer  the  work  moves  in  a  straight  line, 
but  the  cutting  speed  is  still  the  rate  at  which  the  work 
passes  the  tool.    On  the  shaper  the  tool  moves  instead  of 
the  work  and  the  cutting  speed  is  the  rate  at  which  the  tool 
passes  the  work,  while  on  the  milling  machine  the  cutter 
revolves  and  its  surface  speed  is  also  the  cutting  speed. 

97.  Regulation  of  the  cutting  speed.    The  cutting  speed 
depends  on  four  things  :  (1)  the  kind  of  material  to  be  cut ; 
(2)  the  feed  of  the  tool ;  (3)  the  depth  of  cut ;  and  (4)  the 
kind  of  tool  used. 

In  solving  the  problems  that  follow,  the  pupil  must  select  certain 
speeds  and  feeds  as  given  in  Tables  I  and  II.  The  roughing  cut  or 
cuts  are  the  heavy  cuts  taken  to  reduce  the  size  of  work  to  within 
fa"  or  J/'  of  the  correct  size.  The  finishing  cut  is  the  cut  that  reduces 
the  work  to  the  exact  size  required  ;  it  is  seldom  more  than  fa"  deep. 

TABLE  I.    CUTTING  SPEEDS 


MATERIAL 

REVOLUTIONS  PER  MINUTE 

Lathe 

Planer 

Miller 

Drill 

Steel  or  wrought  iron 
Cast  iron  .... 
Tool  steel  .... 
Brass 

25  to    45 
40  to    60 
20  to    40 
100  to  120 
60  to    80 

20  to  30 
25  to  30 

25  to    30 
40  to    60 
20  to    40 
100  to  120 

30  to    40 
40  to    60 
20  to    30 
100  to  120 

Phosphor  bronze 



119 


i<-  N) 

»  **  •$  ^    ^S' 

i  h-H^ 

HiJ.^ 


120 


CUTTING  SPEED  AND  FEED 


121 


In  the  above  table  the  higher  speeds  may  be  used  for  the  lighter 
cuts  and  smaller  diameters. 

The  cutting  speeds  given  here  are  for  carbon  steel.  For  "  high- 
speed "  tools  the  cutting  speed  and  feed  may  be  nearly  doubled.  For 
drills  the  peripheral  speed  is  assumed  as  the  cutting  speed. 

TABLE  II.    FEEDS 


MATERIAL 

FEED  IN  INCHES  PER  REVOLUTION 

Lathe 

Planer 

Miller 

Drill 

Steel  or  wrought  iron 
Cast  iron  .... 
Tool  steel  .... 
Brass  

TV  to  & 
T1*  tO  ^ 
aV  tO  Jo 

1-V  to  A 

sV  to  i 
iV  to  \ 

.008  to  .028 
.008  to  .040 
.008  to  .020 
.010  to  .040 

.005  to  .015 
.008  to  .015 
.005  to  .010 
010 

98.  Feed.    There  are  several  ways  of  expressing  the  feed. 

(1)  The  feed  is  the  distance  in  inches  that  the  tool  ad- 
vances along  the  work  at  every  revolution  or  stroke. 

(2)  Feed  is  sometimes    given   as   the  number  of  turns 
which  the  work  makes  while  the  tool  advances  1  inch,  as 
"  50  or  60  turns  to  the  inch."    This  is  also  equivalent  to 
-^G"  or  ^L"  per  revolution. 

(3)  Feed  may  be  given  in  inches  per  minute,  as  is  often 
the  case  in  connection  with  the  milling  machine.    This  last 
method  is  very  convenient  in  estimating  the  time  necessary 
to  finish  a  given  cut.    The  feed  when  thus  given  can  also  be 
expressed  as  in  (1)  if  the  R.  P.  M.  are  known. 

99.  Rule  for  feed.    A  simple  rule  for  feed  is  :   The  feed  is 
the  distance  moved  by  the  tool  divided  by  the  number  of 
revolutions  made  by  the  work  while  the  tool  is  moving  that 
distance. 

For  convenience,  the  distance  moved  is  generally  taken 
as  1 ".  For  example,  a  lathe  spindle  revolves  75  times  while 
the  carriage  moves  1".  Then  the  feed  is  7y.  Or,  a  milling 
cutter  revolves  65  times  while  the  table  moves  1".  Then 


122  SHOP  PROBLEMS  IN  MATHEMATICS 

the  feed  is  ffy.    Or,  the  shaper  makes  35  strokes  while  the 
table  (that  holds  the  work)  moves  1".    Then  the  feed  is  ^". 

PROBLEMS 

1.  Denoting  the  feed  by  F,  the  number  of  revolutions  by 
N,  and  the  distance  that  the  tool  moves  .by  D,  write  the 
formula  for  F  in  terms  of  D  and  N. 

2.  Solve  this  formula  for  D  and  N. 

3.  A  lathe  tool  moves  2£"  along  the  work  in  1  min.  and 
the  speed  of  the  lathe  is  400  E.  P.  M.    Find  the  feed. 

4.  A  piece  of  work  revolves  60  times  while  the  tool  moves 
f".    Find  the  feed. 

5.  A  planer  makes  35  strokes  while  the  tool  moves  1-J-". 
Find  the  feed. 

6.  If  a  milling  cutter  revolves  100  times  while  the  table 
moves  1",  what  is  the  feed  ? 

7.  A  milling  cutter  has  a  feed  of  .017"  per  revolution 
and  turns  at  160  E.  P.  M.    What  is  the  feed  per  minute  ? 

8.  A  high-speed  drill  f  "  in  diameter  drills  through  237 
pieces  of  cast  iron  3£"  thick  without  being  resharpened. 
A  carbon  drill  of  the  same  diameter  drills  only  55  pieces. 
How  many  feet  in  length  does  each  drill  cut  ? 

9.  By  what  per  cent  is  the  high-speed  drill  better  than 
the  carbon  drill  ? 

10.  How  many  pounds  of  iron  does  each  of  the  above 
drills  remove? 

11.  A  high-speed  drill  of  $•"  diameter  revolving  at  250  revo- 
lutions per  minute  may  be  fed  TJ^"  per  revolution.    How 
long  would  it  take  to  drill  through  a  piece  of  iron  1J"  thick  ? 

12.  If  the  proper  feed  per  revolution  for  a  if"  drill  is 
.008",  how  many  revolutions  will  the  drill  make  in  passing 
through  a  piece  of  iron  3T^"  thick? 


CUTTING  SPEED  AND  FEED  123 

13.  At  a  speed  of  268  revolutions  per  minute,  how  long 
will  it  take  to  drill  through  the  piece  of  cast  iron  mentioned 
in  Problem  12  ? 

14.  A  high-speed  drill  of  f "  diameter  is  required  to  drill 
f "  pieces  of  cast  iron  in  28  sec.    What  feed  and  how  many 
revolutions  per  minute  would  you  have  to  give  the  drill? 

Allow  a  cutting  speed  of  60  ft.  a  minute  for  drills. 

15.  A  cylinder  3  ft.  10"  long  and  9"  in  diameter  is  turned 
to  81"  in  a  lathe  with  a  cutting  speed  of  25  ft.  a  minute  and 
a  feed  of  TL".    How  long  will  it  take  to  make  the  roughing 


FIG.  74.    SINGLE-THROW  CRANK 

cut  ?  How  long  will  it  take  to  make  the  finishing  cut,  using 
a  feed  of  ±"? 

16.  A  boring  bar  is  to  be  35"  long  and  If"  in  diameter. 
Allow  a  cutting  speed  of  50  ft.  and  feeds  of  ¥y  and  ^\" 
respectively  for  roughing  and  finishing.    Find  the  time  re- 
quired to  complete  the  work,  using  If"  stock. 

17.  A  lot  of  16  spindles,  15 £"  long,  are  to  be  turned  down 
from  2"  to  1£"  in  diameter  for  15"  of  their  length.    How 
long  will  it  take  with  a  cutting  speed  of  40  ft.,  a  feed  of 
Jj"  (roughing)  and  3y  (finishing),  and  a  depth  of  cut  not 
to  exceed  ^"? 

18.  Estimate  the  time  necessary  to  finish  the  crank  shown 
in  Fig.  74.    Owing  to  the  opening  between  the  arms,  either 
a  low  cutting  speed  or  a  light  cut  must  be  used. 

19.  A  taper  ring-mandrel  £"  at  the  small  end  and  1^" 
at  the  large  end  is  turned  from  a  If"  round  bar.    Estimate 


124 


SHOP  PROBLEMS  IN  MATHEMATICS 


the  time  required  to  do  the  work  if  the  surface  speed  em- 
ployed is  40  ft.  a  minute,  the  feed  ¥y,  and  the  depth  of 
cut  does  not  exceed  ^". 

20.  Estimate  the  time  required  to  make  the  phosphor- 
bronze  bearing  shown  in  Fig.  75.    Time  must  be  allowed  for 
setting  the  taper  attachment  and  setting  for"  the  thread. 

21.  If  a  milling  cutter  revolves  85  times  per  minute  with 
a  table  feed  of  6Jg  "  per  minute,  what  is  the  feed  per  revolu- 
tion ?  What  is  the  feed  per  tooth  with  a  10-tooth  cutter  ? 

22.  A  cutter  revolves  at  203  R.  P.  M.  with  a  table  feed  of 
9^"  per  minute.    Find  the  cutting  speed  for  a  f"  end  mill 

r/MiMrtfi __        with  10  teeth.   Find 
UF   ::::::miflO 


<o|<0 


the  feed  per  tooth. 

23.  A  cutter  is  3" 
in  diameter  and  4" 
long.  The  depth  of 
cut  in  milling  a  piece 

of  cast  iron  is  T\", 
FIG.  75.    BEARING  the  feed  3,,  &  ^^ 

and  the  cutting  speed  25  ft.  a  minute.    Find  the  weight 
of  metal  removed  per  hour. 

24.  If  an  end  mill  li"  in  diameter  has  12  teeth  and  is 
to  mill  a  surface  4"  x  16",  using  the  end  of  the  mill,  find 
the  R.  P.  M.  and  the  total  time  required.    Assume  a  cutting 
speed  of  40  ft.  per  minute  and  a  feed  of  .004"  per  tooth. 

25.  An  end  mill  is  -J-"  in  diameter  and  has  6  teeth.    Find 
the  number  of  revolutions  necessary  to  give  the  mill  a  sur- 
face speed  of  40  ft.  a  minute. 

26.  Find  the  feed  per  minute  in  Problem  25,  allowing  a 
chip  .005"  thick  for  every  tooth. 

27.  A  milling  cutter  3"  in  diameter  has  24  teeth,  and  is 
fed  into  the  work  .117"  per  revolution.    How  much  of  a  cut 
is  every  tooth  taking  ? 


CUTTING  SPEED  AND  FEED  125 

28.  A  milling  cutter  is  3y  in  diameter  and  has  16  teeth. 
It  revolves  at  52  R.  P.  M.  and  the  table  travels  5T£"  per 
minute.    Find  the  thickness  of  cut  for  every  tooth. 

29.  A  milling  cutter  is  2^-"  in  diameter  and  has  15  teeth. 
If  it  has  a  cutting  speed  of  60  ft.  per  minute,  find  the 
feed  per  minute  when  every  tooth  is  allowed  to  cut  a  chip 
.003"  thick. 

30.  A3"  milling  cutter  may  have  a  cutting  speed  of  38  ft. 
per  minute,  a  depth  of  cut  T^",  and  a  feed  of  .005"  per  tooth. 
Find  the  number  of  revolutions  of  the  spindle  per  minute,  the 
feed  per  minute,  and  the  weight  of  steel  removed  per  minute. 


Anqular 


FIG.  76.    CROSS  SLIDE 


16 


31.  An  angle  iron  whose  sides  are  each  8"  square  is  to 
be  milled  on  the  two  outside  surfaces  with  a  1^-"  end  mill. 
Two  cuts  must  be  taken,  a  roughing  and  a  finishing  cut.    If 
the  cutting  speed  is  40  ft.  a  minute  and  the  feed  3^-"  a 
minute,  find  the  time  required. 

32.  A  cross  slide  (Fig.  76)  is  to  be  milled  on  the  bottom 
with  a  2-j-"  plain  mill  4^"  long.    Find  the  time  required  with 
a  cutting  speed  of  40  ft.  a  minute  and  a  feed  of  .050"  per 
revolution  of  the  cutter. 

33.  The  surfaces  A  and  B  (Fig.  76)  are  to  be  milled  first 
with  a  gang  mill,  as  shown  by  the  dotted  lines,  and  second 
by  an  angular  cutter  held  in  a  vertical  attachment.    How 
long  will  it  take  to  make  the  cut  with  the  gang  mill  by  using 
a  feed  of  .063"  per  revolution  and  a  speed  of  47  E.  P.  M. 


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FIG.  77  J5.    EXERCISES  IN  MACHINE  WORK 
127 


128  SHOP  PROBLEMS  IN  MATHEMATICS 

34.  With  a  cutting  speed  of  35  ft.  a  minute  for  the  angu- 
lar cutter  and  feeds  of  .050"  and  .032"  for  roughing  and 
finishing  cuts  respectively,  find  the  total  time  for  the  4  cuts 
(2  on  each  side)  for  the  angle. 

35.  Find  the  total  time  for  completing  the  cross  slide, 
allowing  2  min.  for  the  setting  of  the  work' on  the  milling- 
machine  table. 

36.  Estimate  the  time  required  to  finish  the  pieces  shown 
in  Fig.  77.    These  are  to  be  made  in  lots  of  50. 

The  time  required  to  set  up  the  machines  ready  for  the  various 
operations  should  be  divided  by  50  and  added  to  the  time  necessary 
to  do  the  actual  machining.  Suitable  cutting  speeds  and  feeds  may  be 
selected  from  the  preceding  tables.  Those  exercises  requiring  threads 
to  be  cut  on  the  lathe  should  be  omitted  until  the  chapter  on  Thread 
Cutting  has  been  taken  up.  With  the  exception  of  the  pulley,  the 
stuffing  box,  and  the  right  and  left  coupling,  the  pieces  are  all  made 
of  machine  steel. 

100.  Cutting  speeds  on  the  planer  and  shaper.  The  calcula- 
tions of  cutting  speeds  on  the  planer  and  shaper  are  some- 
what more  difficult  than  on  other  machines  on  account  of 
their  forward  and  reverse  movements,  arid  because  the  re- 
verse is  generally  faster  than  the  forward  stroke. 

Henceforth  the  term  "stroke  "  will  mean  the  complete  forward  and 
backward  movement,  except  when  the  reference  is  to  the  length  of 
the  stroke. 

EXAMPLE.  If  a  shaper  makes  22  strokes  in  1  min.,  the 
reverse  is  2  to  1  of  the  forward  stroke,  and  the  length  of 
the  stroke  is  1  ft.,  what  is  the  cutting  speed  ? 

If  the  length  of  stroke  is  1  ft.,  then  the  total  distance 
traveled  in  one  direction  in  1  min.  is  22  x  1  ft.,  or  22  ft.  But 
only  f  of  the  time  is  used  in  the  forward  movement,  so  that 
the  cutting  speed  per  minute  is  22  ft.  x  f ,  or  33  ft.  (In  other 
words,  if  the  shaper  travels  22  ft.  in  f  min.,  in  1  min.  it 
would  travel  22  ft.  x  },  or  33  ft.) 


CUTTING  SPEED  AND  FEED  129 

Now  suppose  the  length  of  the  stroke  is  \  ft.  instead  of 
1  ft.  The  total  distance  traveled  is  22  x  i  ft.,  or  11  ft.,  in 
|  of  1  min.,  or  at  the  rate  of  16^  ft.  per  minute. 

It  is  evident  that  to  obtain  the  same  cutting  speed  for 
all  lengths  of  stroke  the  E.  P.  M.  should  be  increased  for 
shorter  and  decreased  for  longer  strokes. 

The  cutting  speed  of  the  shaper  varies  with  the  length  of  the 
stroke  ;  on  the  other  hand,  the  length  of  stroke  does  not  depend  upon 
the  number  of  strokes  per  minute.  In  the  case  of  the  planer  the 
longer  its  stroke,  the  fewer  strokes  it  will  make  per  minute,  but  the 
cutting  speed  is  constant,  no  matter  what  the  length  of  the  stroke 
may  be. 

EXAMPLE.  A  planer  makes  12  strokes  per  minute,  2  ft. 
in  length,  and  the  reverse  is  2  to  1.  Find  the  cutting  speed. 

The  total  distance  traveled  in  one  direction  is  12  x  2  ft., 
or  24  ft.  The  actual  time  taken  in  moving  this  distance 
is  f  min.  Then  the  cutting  speed  is  24  ft.  x  f ,  or  36  ft.  per 
minute. 

PROBLEMS 

1.  A  shaper  makes  36  strokes  per  minute  and  has  a  2  to 
1  reverse.    What  is  its  cutting  speed  with  a  16"  stroke  ? 
10"  stroke?  2"  stroke? 

2.  Ascertain  the  number  of  strokes  per  minute  that  would 
be  necessary  to  give  a  cutting  speed  of  36  ft.  per  minute 
for  each  of  the  following  lengths  of  stroke  of  the  above 
shaper:  2",  6",  10",  14". 

3.  Estimate  the  time  necessary  to  finish  upon  the  shaper 
the  slide  shown  in  Fig.  76,  considering  that  it  can  be  done 
in  two  cuts,  roughing  and  finishing,  allowing  1^  min.  for 
setting  the  head  for  the  dovetail  and  1  min.  for  turning 
over.    Allow  a  cutting  speed  of  30  ft.  per  minute. 

4.  A  planer  makes  10  strokes  per  minute ;  the  length  of  the 
stroke  is  4  ft.  and  the  reverse  2  to  1.    Find  the  cutting  speed. 


130 


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132  SHOP  PROBLEMS  IN  MATHEMATICS 

5.  A  planer  is  constructed  so  as  to  give  a  cutting  speed 
of  50  ft.  per  minute  and  a  reverse  speed  of  70  ft.  per  minute. 
Find  the  number  of  strokes  that  it  would  make  for  the  fol- 
lowing lengths  of  strokes :  3  ft.,  4  ft.,  5  ft.,  6  ft. 

6.  A  planer  has  a  cutting  speed  of  25  ft.  per  minute  and 
a  reverse  of  2  to  1.    Find  the  number  of  strokes  per  minute 
for  the  following  lengths  of  stroke :  6",  2  ft.,  4  ft.,  iO  ft. 

7.  A  planer  having  an  8-ft.  table  can  hold  five  cross  slides 
(Fig.  76)  at  one  time.    Allowing  a  stroke  of  7^  ft.,  a  cutting 
speed  of  30  ft.  per  minute,  and  a  reverse  of  50  ft.  per  minute, 
estimate  the  time  (per  piece)  required  to  finish  the  work. 
Allow  1^  min.  for  setting  the  head  for  the  dovetails  and 
2  min.  for  turning  over  preparatory  to  planing  the  bottoms. 

8.  Compare  the  results  found  for  this  slide  when  done 
upon  the  miller  (Problem  35,  sect.  99);  the  shaper  (Prob- 
lem 3) ;  the  planer  (Problem  7). 

9.  A  casting  4  x  10  ft.  is  to  be  finished  on  a  planer.   The 
cutting  speed  is  20  ft.  a  minute  and  the  feed  J-"  for  every 
stroke.    Find  the  time  required  to  take  one  complete  cut. 

The  reverse  stroke  may  be  considered  as  taking  one  half  the  time 
of  the  forward  stroke. 

10.  A  planer  with  a  7  x  3  ft.  bed  moves  at  the  rate  of 
40  ft.  a  minute  on  the  cutting  stroke  and  60  ft.  a  minute 
on  the  back  stroke.   Estimate  the  time  necessary  to  finish  a 
series  of  six  surface  plates  2  x  3  ft.    The  feed  for  roughing 
may  be  Ty  and  for  finishing  T3g-". 

11.  The  planer  makes  21  strokes  in  1  min.  and  the  length 
of  travel  is  27".    What  is  the  surface  speed  forward  and 
backward,  if  the  ratio  of  forward  to  reverse  drive  is  1  to  3, 
and  no  time  is  allowed  for  reversing  ? 

12.  What  would  be  the  result  in  Problem  11  if  T\y  of  the 
total  time  were  allowed  for  reversing  the  motion  of  the 
planer  table? 


CHAPTER  IX 


MICROMETER,  VERNIER,  AND  TAPERS 
MICROMETER 

For  a  knowledge  of  the  method  of  changing  common  fractions  to 
decimals,  see  section  162. 

101.  Description.  The  names  of  the  different  parts  of  the 
micrometer  are  given  in  Fig.  79.  On  the  spindle  threads  are 
cut  40  to  the  inch. 
This  spindle  is  se- 
cured to  thimble 
and  turns  with  it, 
so  that  one  com- 
plete turn  of  the 
thimble  means  one 
turn  of  the  screw 
and  a  spindle  move- 

1.000 


Anvil 


Spindle 


Barrel 


FIG.  79.    MICROMETER  CALIPERS 


ment  of  ^  of  one  inch, 


40 


(or  .025  =  twenty-five  thou- 


sandths of  an  inch).  The  spindle  is  made  to  advance  by  a 
nut  in  the  end  of  barrel,  which  may  also  be  adjusted  for 
wear,  for  it  is  evident  that  there  must  not  be  any  looseness 
between  the  screw  and  the  nut.  The  barrel  is  generally  di- 
vided for  the  distance  of  an  inch  into  ten  equal  parts,  tenths 
(y^or.l)  (sometimes,  however,  into  eighths),  and  each  T^  into 
4  parts,  making  a  total  of  40  divisions  on  the  barrel,  the  same 
number  as  the  threads  per  inch.  Hence  one  small  division  on 
the  barrel  is  equal  to  one  turn  of  the  screw,  which  is  equal 
to  ¥y,  or  .025".  Four  small  divisions  on  the  barrel  or  one 

133 


134  SHOP  PROBLEMS  IN  MATHEMATICS 

large  one  =  4  x  .025"  =  .100"  =  Ty '.  It  will  be  seen  also  that 
the  edge  of  the  thimble  is  divided  into  25  parts.  Therefore 
since  one  complete  turn  of  the  thimble  is  equal  to  .025",  or 
JQ  ",  one  division  on  the  thimble  will  be  equal  to  ^  of  .025", 
or  .001"  GV  of  ?V  =  TO'OO  =  -001). 

Hence    1  division  on  the  thimble  is  equal  to  .001". 

1  small  division  on  the  barrel  is  equal  to  .025". 
1  large  division  (4  small  ones)  is  equal  to  .100". 
10  large  divisions  are  equal  to  1.000". 

102.  How  to  read  the  micrometer.  When  it  is  desired  to  set 
any  decimal  upon  the  micrometer,  divide  the  number  into 
groups  of  tenths  (large  divisions),  .025"  (small  divisions), 
and  .001"  (thimble  divisions).  For  example,  .637"  is  equal 
to  6  large  divisions,  1  small  division,  and  12  thimble  divi- 
sions (.6" +  .025" +.012"  =  .637");  or,  as  we  read  from 
the  micrometer  a  new  reading,  .7"  +  .050"  +  .017"  =  .767", 
which  is  7  large  divisions,  2  small  divisions,  and  17  divisions 
on  the  thimble. 

EXAMPLE.    Divide  .397"  into  its  proper  divisions. 

.397"  =.3" +  .075" +  .022". 

PROBLEMS 

1.  In  measuring  a  piece  of  work  with  the  micrometer  the 
following  readings  are  made,  each  by  a  different  student : 
1.3785",  1.3783",  1.3779",  1.3781",  1.3778",  1.3785".  What 
is  the  probable  diameter  and  what  the  greatest  error  ? 

2.  When  the  jaws  of  a  micrometer  are  closed  the  reading 
is  .0003.    Should  this  reading  be  added  to  or  subtracted  from 
readings  in  order  to  find  the  true  reading  ? 

3.  Divide  the  following  decimals  into  their  correct  group- 
ings for  the  micrometer  by  the  method  shown  in  section  102  : 
.149",  .217",  .393",  .403",  .4875",  .6375",  .741",  .828",  .967", 
.9995". 


MICROMETER,  VERNIER,  AND  TAPERS         135 

4.  Change  the  following  fractions  to  decimals  of  an  inch 
and  group  them  for  setting  on  the  micrometer:  £",  Ty,  1", 

5    II     3  II       7    II       9    II     5>l      11"     25"     27" 
TF    >    t    >    16     »!¥>¥>    1*    '   3*    >    64     ' 

5.  With  the  micrometer  of  Problem  2  a  piece  of  work 
measures  0.3895".    What  is  its  true  diameter? 

6.  Apiece  of  work  is  to  be  finished  exactly  .6375".   What 
reading  should  be  made,  if  the  micrometer  when  closed  reads 
.0004"  less  than  0  ? 

7.  A  piece  of  steel  which  is  to  measure  .995"  is  heated 
to  215°  F.  while  grinding  and  then  calipers  .996".  What  will 
it  measure  at  65°  F.?  (Coefficient  of  expansion  =  .0000066" 
for  each  degree.) 

8.  A  space  between  two  finished  surfaces  of  iron  was  found 
to  be  filled  exactly  by  5  sizing  pieces  measuring  respectively 
.375",  .050",  .020",  .010",  .002".    How  large  was  the  space  ? 

A  B 


FIG.  80.    SECTION  OK  MICROMETER  CALIPERS 

9.  One  surface  of  a  jig  must  be  just  .387"  higher  than 
another.    This  surface  is  to  be  measured  by  piling  up  siz- 
ing blocks  to  the  distance  required  and  then  making  the 
surface  level  with  these  blocks.    Select  the  required  pieces 
from  the  following:  .375",  .150",  .100",  .050",  .020",  .014", 
.012",  .011",  .010",  .005". 

10.  Select  from  the  above  sizing  blocks  the  required  num- 
ber to  give  the  distances  as  follows  :  .437",  .529",  .233",  .137". 

On  account  of  the  size  of  the  divisions  on  the  thimble  D  it  becomes 
easy  to  estimate  to  one  quarter  of  a  thousandth  (.00025")  with  only 
a  slight  error.  For  example,  in  Fig.  80,  A ,  the  reading  is  nearly  .4065", 
and  in  Fig.  80,  B,  it  is  nearly  .4348". 


136  SHOP  PROBLEMS  IN  MATHEMATICS 

Some  micrometers  have  an  additional  set  of  lines  on  the  barrel, 
called  a  "vernier"  (see  following  section),  for  accurate  reading  to  ten 
thousandths.  Since  holding  the  micrometers  in  the  hand,  as  the  work- 
man usually  does,  may  cause  them  to  vary  the  readings  by  at  least 
.0002",  the  use  of  the  vernier  is  of  value  only  with  the  most  delicate 
handling. 

THE  VERNIER 

103.  Vernier.    The  vernier  is  an  instrument  for  measuring 
very  small  distances.    It  was  named  for  its  inventor,  Pierre 
Vernier,  who  first  used  it  in  1631. 

104.  The  vernier  of  the  micrometer.    A  micrometer  reading 
to  thousandths  of  an  inch   may  be  made  to  read  to  ten 
thousandths  of  an  inch  by  the  addition  of  a  vernier  on  the 

i ~  c  A   barrel.     Ten  divisions  on   the 

vernier  correspond  to  nine  di- 
visions on  the  thimble.  There- 
fore the  difference  between  the 

.xx  c  /\  width  of  one  division  on  the  ver- 

nier and  one  division  on  the 

FIG.  81.    VERNIER  ,,  .     ,  ,  ,-,        ,,          -,. 

thimble  is  one  tenth  of  a  di- 
vision on  the  thimble,  or  one  ten-thousandth  (.0001)  of  an 
inch.  For  example,  in  Fig.  81  the  third  line  from  0  on  the 
thimble  coincides  with  the  zero  line  of  the  vernier.  The 
next  line  on  the  thimble  is  distant  from  the  next  line  of 
the  vernier  by  one  tenth  of  a  division  on  the  thimble.  The 
next  two  lines  on  the  thimble  and  vernier  respectively  are 
separated  by  two  tenths  of  a  thimble  space,  the  next  two 
by  three  tenths,  etc. 

In  opening  the  micrometer  every  space  on  the  thimble 
represents  an  opening  of  one  thousandth  of  an  inch.  If  the 
thimble  is  turned  to  the  left,  so  that  the  first  division  line 
on  the  vernier  coincides  with  the  fourth  division  line  of  the 
thimble,  the  micrometer  has  been  opened  one  tenth  of  one 
thousandth,  or  one  ten-thousandth,  of  an  inch.  If  the  second 


MICROMETER,  VERNIER,  AND  TAPERS         137 

division  line  of  the  vernier  coincides  with  the  line  marked 
5  on  the  thimble,  the  tool  has  been  opened  two  ten-thou- 
sandths, etc. 

Hence,  to  read  the  vernier  of  a  ten-thousandths  microm- 
eter, observe  the  line  on  the  barrel  that  coincides  with  a 
line  on  the  thimble.  If  the  line  is 
marked  1,  the  reading  is  one  ten- 
thousandth  ;  if  the  line  is  marked 
2,  the  reading  is  two  ten-thou- 


sandths, etc. 

T7<  T         T7<'  00       F  4-T,  FlG .    82.       VERNIER 

EXAMPLE.   In  Fig.  82,  6  on  the 

vernier  coincides  with  a  division  line  on  the  thimble.  The 
vernier  reading,  therefore,  is  .0006".  If  the  reading  of  the 
micrometer  to  thousandths  is  .325",  the  complete  reading 
is  .3256". 

TAPERS 

105.  Taper.   A  piece  of  machine  work  is  said  to  be  tapered 
when  it  gradually  increases  in  diameter  or  thickness.    It 
may  be  round  as  in  a  lathe  center,  or  flat  as  in  a  taper  gib. 
The  amount  of  taper  can  be  given  only  when  the  increase 
is  uniform  (see  Fig.  83). 

106.  Method  of  expressing  tapers.    Taper  is  most   com- 
monly given  in  inches  per  foot.    It  is  the  difference  be- 
tween the  diameters  (in  inches)  of  the  ends  of  a  piece  of 
work  1  ft.  long.    To  ascertain  the  amount  per  foot  when 
the  taper  is  actually  more  or  less  than  a  foot  in  length  the 
difference  in  diameter  for  1  ft.  must  be  calculated.    Flat 
tapered  work  may  be  given  in  the  same  way. 

Generally  when  the  taper  is  excessive  its  value  is  given 
in  degrees,  as  in  the  case  of  bevel  gears,  lathe  centers,  etc. 
(see  Fig.  88). 

The  taper  is  also  sometimes  expressed  in  fractions  of  an 
inch  per  inch  of  length,  as  T^"  to  1". 


138 


SHOP  PROBLEMS  IN  MATHEMATICS 


This  method  is  often  more  convenient  for  the  workman, 
saving  him  the  trouble  of  changing  the  amount  of  the  taper 
per  foot  to  the  per-inch  basis. 

107.  Length  of  tapers.  In  turning  tapers,  if  the  tail- 
stock  is  to  be  offset,  then  the  actual  length  of  the  work  or 
the  mandrel,  and  not  merely  the  length  of  the  tapered  part 

T — r of  the  work,  must  be 

-  '  A 


considered  in  the  cal- 
culations. If  the  taper 
attachment  is  to  be 
used,  it  is  only  neces- 
sary to  find  the  correct 


— » 


B 

"jy 

J_ 

*                   </l"                     >- 

_  

<                       3"    •     -•    —  * 

•^    " 

FIG.  83.    TAPERS 

taper  in  inches  per  foot  and  set  the  taper  attachment  to  that. 
No  attention  need  be  paid  to  length  of  work  or  mandrel. 

108.  Method  of  calculating  tapers. 

(a)  Where  the  tailstock  is  to  be  offset.  CASE  1.  In  piece 
A,  shown  in  Pig.  83,  the  difference  in  diameter  between 
the  large  and  the  small  ends  is  If"  —  1"  =  f ".  Take  one 
half  of  this  difference  for  the  offset  of  the  tailstock,  or 
i  of  f  "  =  f". 

CASE  2.  In  piece  B  the  taper  does  not  extend  from  end 
to  end,  but  it  must  be  treated  as  though  it  did  extend  the 


MICROMETER,  VERNIER,  AND  TAPERS         139 

whole  length  and  the  difference  in  diameter  at  the  ends 
calculated.  The  difference  in  diameter  in  3"  of  length  is 
1|"  _  1"  =  f .".  In  1"  of  length  the  difference  in  diameter 
would  be  ^  of  f  "  =  y\".  If  the  piece  is  10"  long,  then 
10  x  fV  =  2i",  which  is  the  total  difference  in  diameter 
at  the  ends.  If  the  piece  were  7£"  long,  then  1\  x  TV  =  If  V- 
For  the  offset  one  half  of  2^-"  must  be  taken. 

(b)  When  the  taper  attachment  is  to  be  used.  Eem ember 
that  in  setting  the  taper  attachment  the  taper  per  foot 
must  be  obtained.  In  A  (see  Fig.  83)  the  total  difference  in 
diameters  is  f ";  then  the  taper  per  inch  is  -J  of  f  ",  or  £"  per 
inch.  In  12"  it  would  be  12  x  i",  or  3"  per  foot.  In  B  the 
taper  per  inch  was  found  to  be  ^\";  then  the  taper  per  foot 
is  12  x  f\ "  —  3",  no  matter  what  is  the  length  of  the  whole 
piece.  Having  found  the  taper  per  foot,  it  is  only  necessary 
to  set  the  taper  attachment  to  that  amount. 

PROBLEMS 

1.  Make  a  simple  formula  for  the  offset  of  the  tailstock 
in  taper  turning,  letting  D  equal  the  large  diameter,  d  the 
small  diameter,  and  s  the  offset. 

2.  A'  lathe  center  is  TV  in  diameter  at  the  small  end. 
If  the  taper  were  |"  per  foot,  what  would  be  the  diameter 
4"  from  the  small  end  ? 

3.  Change  the  following  from  taper  per  inch  to  taper  per 
foot :  .0833",  .0625",  ,052",  .0416",  .0312",  .0208",  .0104". 

4.  A  lathe  center  is  1.02"  in  diameter  at  the  small  end. 
If  the  taper  per  foot  is  .623",  what  is  the  diameter  4"  from 
the  end  ? 

5.  What  difference  in  diameter  would  there  be  at  the 
large  end  of  two  pieces  4"  long,  if  each  was  Ty  at  the  small 
end  and  one  had  a  taper  of  £"  to  the  foot,  the  other  .602" 
to  the  foot  ? 


140 


SHOP  PROBLEMS  IN  MATHEMATICS 


o| 
"r\J 

\ 

:-T 

L 

2. 

._*__ 



—  2"  —  * 

-1 

6.  How  much  larger  would  one  of  two  pieces  be  if  both 
were  f  "  at  the  small  end,  5"  long,  and  one  were  cut  to  a 
|"  taper,  the  other  to  a  Morse  taper  No.  2  ? 

Morse  taper  No.  2  is  .602"  per  foot. 

7.  A  crank  for  a  gas  engine  is  25"  long  and  is  to  have  a 
taper  of  f "  to  the  foot  turned  at  one  end.    How  much  must 
the  tailstock  be  offset  ? 

8.  A  piston  rod  has  a  taper  5"  long,  cut  at  one  end.    At 
the  small  end  of  the  taper  the  rod  measures  1.792"  and  the 

large  diameter  of  the  piston  rod  is 
2".    What  is  the  taper  per  foot  ? 

9.  The  tailstock  of  a  lathe  can 
be  offset  a  distance  of  2|".    What 
is  the  largest  taper  that   can  be 
cut  on  a  bar  10"  long?  2  ft.  long? 
3  ft.  long  ? 

10.  What  is  the  greatest  length 
that  the  piston  rod  in  Problem  8  could  be  and  still  have 
its  taper  cut  by  a  lathe  with  a  tailstock  which  can  be  offset 
not  more  than  2"  ? 

11.  A  plug  gauge  Ty  in  diameter  enters  a  hole  of  un- 
known taper  a  distance  of  2|".    A  plug  of  T9^"  in  diameter 
enters  2.4".    Find  the  taper  per  foot. 

12.  A  lathe  center  is  \%"  in  diameter  at  the  small  end 
and  3"  from  the  end  it  is  §|".    Find  the  taper  per  foot. 

13.  Stock  for  a  milling-machine  arbor  is  14"  long.    It  is 
to  be  tapered  for  a  length  of  6"  on  one  end  to  •£•"  per  foot. 
How  much  must  the  tailstock  be  offset  ? 

14.  A  lathe  spindle  bearing  as  shown  in  Fig.  75  is  to  be 
turned  on  a  mandrel  8"  long.    Find  the  offset  of  the  tail- 
stock.    If  the  mandrel  were   6"  long,  what  would  be  the 
offset  ? 


FIG.  84.    TAPER  BUSHING 


MICROMETER,  VERNIER,  AND  TAPERS         141 

15.  What  is  the  taper  per  foot  of  the  bearing  shown  in 
Fig.  75  ? 

16.  The  bushing  shown  in  Fig.  84  is  to  be  turned  on  a 
mandrel  7£"  long.    Find  the  offset  of  the  tailstock. 

17.  What  is  the  taper  per  foot  of  the  piece  shown  in 
Fig.  84? 

18.  Stock  for  lathe  centers  has  been  cut  off  5J"  in  length. 
How  much  must  the  tailstock  be  offset  to  cut  a  taper  of 
TY'per  foot? 


1_ 


FIG.  85.    LATHE  SPINDLE 


19.  By  use  of  the  table  of  natural  functions  find  the 
approximate  included  angle  for  the  following  tapers  per 

fnnf  •       7    "       9    "      11"     13"     15" 

)0t  .     T%    ,    T%    ,   T^    ,   Tff    ,    Tff    . 

20.  To  what  included  angles  are  the  sleeves  of  the  fol- 
lowing diameters  equal  ? 


Large  diameter 
Inches 

Small  diameter 
Inches 

Length 
Inches 

2 

1 

2 

3 

1 

3 

3 

2 

3 

3 

2 

1 

4 

1 

2 

4 

1 

u 

4 

1 

i 

21.  Find  the  taper  per  foot  of  the  long  taper  of  the 
lathe  spindle  shown  in  Fig.  85. 


142 


SHOP  PROBLEMS  IN  MATHEMATICS 


22.  Find  the  included  angle  of  the  abrupt  taper  of  the 
lathe  spindle  shown  in  Fig.  85. 

23.  A  pulley  with  a  6"  face  is  £"  larger  in  diameter  at 
the  middle  than  at  the  edges.    Find  the  taper  per  foot. 

24.  Find   the    offset   of   tailstock    when   the   pulley  of 
Problem  23  is  turned  on  a  mandrel  11"  long. 

The  increase  in  diameter  toward  the  middle  of  the  face  of  a 
pulley  is  called  crowning.  It  causes  the  belt  to  keep  in  the  center  of 
the  pulley. 

25.  A  pulley  of  1"  face  is  crowning  -fa".    What  is  the 
taper  per  foot  ?    Find  the  offset  of  the  tailstock  when  the 
pulley  is  turned  on  a  6"  mandrel. 


I      Line    of  L^the  Centers  I _J I 


FIG.  86.    METHODS  OF  GRADUATING  THE  BASES  OF  COMPOUND 
SLIDE  RESTS 

109.  Slide  rests.  The  correct  setting  of  slide  rests  of 
lathes  or  the  crossheads  of  planers  and  shapers  for  cutting 
tapers  or  angles  depends  largely  upon  a  clear  understanding 
of  the  relations  of  angles  one  to  another,  such  as  the  com- 
plement and  supplement  of  an  angle ;  also  that  the  three 
angles  of  a  triangle  are  equal  to  180°,  and  that  the  two 
acute  angles  of  a  right  triangle  are  equal  to  90°. 

The  complement  is  equal  to  90°  minus  the  given  angle. 
The  supplement  is  180°  minus  the  given  angle. 

EXAMPLE.  The  complement  of  33°  is  57° ;  of  45°,  45° ; 
of  69°,  21°.  The  supplement  of  120°  is  60°;  of  175°,  5°; 
of  25°,  155°. 


MICROMETER,  VERNIER,  AND  TAPERS         143 

110.  Setting  of  angle.    The  angle  to  which  a  slide  rest 
must  be  set  in  order  to  turn  a  given  angle  depends  on  two 
things :    (1)   how  the   angle    is   given   in   relation  to  the 
center  line  of  the  lathe ;   (2)  at  what  place  on  the  slide 
rest  the  zero  point  is  taken.    A  slide  rest  is  generally  set 
at  zero  when  it  is  in  line  with  the  cross-feed  screw,  or,  in 
other  words,  at  right  angles  to  the  center  line  of  the  lathe. 
Therefore  : 

If  the  required  angle  is  measured  from  a  line  parallel  to 
the  cross-feed,  the  slide  must  be  set  to  the  same  angle  as 
given  on  the  draiving. 

If  the  included  angle  is  given,  the  slide  rest  must  be  set  to 
the  complement  of  one  half  the  included  angle.  (See  Figs.  86, 
87,  and  88.) 

111.  Graduations  on  slide  rests.    The  diagrams  shown  in 
Fig.  86  will  give  some  idea  of  the  different  ways  in  which 


M   '        N          0          PR         S          T 

FIG.  87.    ANGLES  FOR  COMPOUND  SLIDE  REST 

the  graduations  are  put  upon  the  circular  bases  of  com- 
pound slide  rests.  Notice  that  except  in  E  the  zero  marks 
on  the  upper  and  lower  slides  coincide  when  the  slide  is 
set  at  right  angles  to  the  center  line  of  the  lathe. 

All  angles  which  the  student  may  meet  in  his  work  may 
be  resolved  into  one  of  the  cases  shown  in  Fig.  87.  For 
example,  slides  A,  B,  C,  D  may  be  set  directly  to  the  angle 
M.  Slide  E  must  be  set  to  the  complement  of  M,  or  55°. 
Slide  D  may  be  set  direct  to  angle  N,  but  for  slides  A,  B  a 
new  zero  must  be  made  on  the  lower  slide  at  a,  and  the 
slides  brought  around  until  65°  —  45°,  or  20°,  as  shown  by 
the  arrow,  is  opposite  a.  The  arrow  on  C  shows  the  setting 


144 


SHOP  PROBLEMS  IN  MATHEMATICS 


i 


H 


FIG.  88.    BEVEL  GEARS  AND  LATHE  CENTER 


of  that  slide  for  angle  N.    Slide  E  is  set  at  the  comple- 
ment of  65°,  or  25°,  as  shown  by  the  arrow  on  E. 

Angles  O  and 
P  may  be  set  di- 
rectly on  slide  E, 
bpt  on  slides  A, 
B,  C,  D  their  com- 
plements must  be 
found,  30°  and  75° 
respectively.  The  setting  is  then  made  as  in  M  and  N. 

In  the  case  of  included  angles  the  student  may  find  the  cor- 
rect reading  to  which  to  set  the  slide  rest  by  constructing  an 
auxiliary  triangle,  as  at  F'  (see  Fig.  88).  In  the  case  of  bevel 
gears  the  angles  given  in  G  are  the  required  angles,  while 
those  given  in  H  are  not.  By 
constructing  the  auxiliary 
triangles, as  at  F',  the  correct 
angles  may  be  easily  obtained. 
On  the  shaper  and  planer 
the  slide  is  always  vertical 
when  the  scale  is  at  0.  Then 
the  scale  in  Fig.  89  A  would 
indicate  that  the  slide  was 
set  to  an  angle  of  15°  to  the 
vertical.  It  follows  that 

Whatever  angle  is  given  on 
the  drawing,  the  correspond- 
ing angle  with  the  vertical 
should  be  found  and  the  slide 

set  to  that  angle.  FlG"  S9A'  PLANER  CROSS  SLIDE 

In  using  the  universal  head  on  the  universal  cutter  grinders 
the  determination  of  the  correct  angles  is  often  confusing. 
Then  a  sketch  showing  clearly  the  different  angles  will  help 
much  in  obtaining  a  correct  solution. 


MICROMETER,  VERNIER,  AND  TAPERS         145 
PROBLEMS 

1.  A  triangle  has  two  of  its  angles  45°  and  70°  respec- 
tively.   What  is  the  third  angle  ? 

2.  A  right  triangle  has  one  angle  equal  to  57°.    What  is 
the  value  of  the  third  angle  ? 

3.  A  triangle  has  two  equal  angles  and  an  obtuse  angle 
of  127°.    Find  the  value  of  the  other  angles. 

4.  A  perpendicular  line  is  dropped  from  the  vertex  of 
the  obtuse  angle  (Problem  3)  to  the  opposite  side.    Find 
the  value  of  the  angles  of  the 

two  triangles  thus  made. 

5.  Find    the    angles    to 
which   the    top    slide   of    a 
compound  rest  should  be  set 
in  order  to  turn  the  tapers 
found   in  Problem   20,  sec- 
tion 108. 

6.  A  lathe  tail  stock  can  be 
set  over  either  way  3"  from 

the  center  line.  What  is  the 

FIG.  89  B.    LATHE  CROSS  SLIDE 
greatest   taper  that  can  be 

turned  on  a  piece  of  work  6"  long  ?    8"  ?    10"  ?    12"  ? 

7.  To  what  included  angles  (approximately)  are  these 
tapers  equal  ? 

8.  At  what  angle  should  the  compound  rest  be  set  in 
order  to  turn  tapers  similar  to  those  found  in  Problem  2  ? 

9.  To  what  angle  must  the  compound  rest  be  set  to  turn 
an  included  angle  of  30°  ?  40°  ?  45°  ?  55°  ?  65°  ?  90°  ? 

10.  A  dovetail  which  is  to  be  cut  on  a  planer  makes  60° 
with  the  horizontal.    To  what  angle  should  the  slide  rest 
be  set  ? 


146  SHOP  PROBLEMS  IN  MATHEMATICS 

11.  Give  the  correct  setting  for  angles  R,  S,  T,  and  U  (see 
Fig.  87)  upon  the  graduated  scales  A,  B,  C,  D,  E  (Fig.  86). 

12.  Give  the  correct  angles  for  setting  a  planer  cross- 
head  for  the  angles  shown  in  Fig.  90. 


6QA~        ~~Z£?"  T*W\87°    ^7        VTT         757 


FIG.  90.    DOVETAILS 

13.  Give   the   correct   setting  for    the   angles   shown  in 
Fig.  88. 

14.  To  what  angle  should  the  compound  rest  be  set  in 
order  to  turn  the  abrupt  taper  on  the  lathe  spindle  shown 
in  Fig.  85  ? 


CHAPTER  X 

THREAD  PROPORTIONS,  GEARING  FOR  SCREW  CUTTING, 
INDEXING 

THREAD  PROPORTIONS 

112.  Number  of  threads  per  inch.    A  screw  is   generally 
determined  by  the  diameter  of  its  body  and  the  number 
of  threads  per  inch ;  also  by  the  shape  of  the  thread. 

113.  Pitch.    The  distance  from  the  top  of  one  thread  to 
the  top  of  the  next  is  called  the  pitch.    It  is  the  reciprocal 
of  the  number  of  threads  per  inch. 

114.  Lead.    The  distance  which  a  screw  would  advance 
when  given  one  complete  turn  is  called  the  lead.    In  single 
threads  the  lead  is  equal  to  the  pitch. 

115.  Depth  and  double  depth.    The  perpendicular  distance 
from  the  top  to  the  bottom  of  the  thread  of  a  screw  is 
called  the  depth.    Twice  this  distance  is  the  double  depth. 

116.  Root  diameter.    The  diameter  at  the  bottom  of  the 
thread,  or  the  outside  diameter  minus  the  double  depth,  is 
called  the  root  diameter. 

117.  Calculations  for  depth.    It  is  very  important  to  be 
able  to  find  the  depth  of  any  thread,  for  upon  that  value 
depends  the  cutting  of   all   threads  and  the  size  of   all 
tap  drills. 

The  depth  of  a  V -thread  is  equal  to  the  altitude  of  an 
equilateral  triangle  whose  sides  are  equal  to  the  pitch  of 
the  thread  (see  Fig.  92). 

This  altitude  is  very  easily  found  by  geometry  or 
trigonometry.  Considering  1"  as  a  standard  or  basis  of 

147 


FIG.  91.    ENGINE  LATHE 
(By  permission  of  The  Hendey  Machine  Co.) 


1.  Lathe  bed. 

2.  Shears  or  ways. 

3.  Lead  screw  and  feed  rod. 

4.  Reverse  rod.* 

5.  Gear  box  and  rocker  handle. 

6.  Gear  box  and  rocker  handle. 

7.  Stop  rod  and  dogs.* 

8.  Apron. 

9.  Carriage. 

10.  Handwheel  for  longitudinal 

feed.    Power-feed   stud   in 
center. 

11.  Power  cross-feed  clutch. 

12.  Lead  screw  half  nut. 

13.  Cross-feed    screw    and    ball 

crank. 

14.  Bridge  of  carriage. 

15.  Cross  slide  of  carriage. 

16.  Clamping  screw  to  hold  car- 

riage. 


17.  Tool  post. 

18.  Ring  and  wedge. 

19.  Top  slide  of  compound  rest. 

20.  Screw   stop  for  thread   cut- 

ting.* 

21.  Swivel  seat  of  compound  rest. 

22.  Cone  pulley. 

23.  Bearings. 

24.  Back  gears. 

25.  Headstock. 

26.  Small  face  plate. 

27.  Live  spindle  and  center. 

28.  Fore  gear  and  lock  pin. 

29.  Tailstock. 

30.  Tailstock  saddle. 

31.  Handwheel  for  tail  spindle. 

32.  Tailstock  clamp. 

33.  Tail-spindle  binder. 

34.  Tail  spindle  and  center. 

35.  Rack  for  moving  carriage. 


*  Parts  marked  with  star  are  peculiar  to  this  lathe. 
148 


THREAD  PROPORTIONS 


149 


measurement,  we  may  find  a  value,  C  (really  the  depth  of 
a  thread  of  1"  pitch),  such  that  the  following  rule  may 
be  derived : 

For  a  screw  of  any  pitch  the  depth  of  thread  is  the  same 
fractional  part  of  C  as  the  given  pitch  is  of  the  base  pitch  1". 


FIG.  92.    V-THREAD 

EXAMPLE.  Find  the  depth  of  a  thread  of  -J"  pitch  whose 
angle  is  55°. 

By  trigonometry  the  altitude  a  of  the  triangle  A  (Fig.  92), 
whose  side  P  is  equal  to  1",  is  found  thus : 


and 


a  = 


.5 


.5206 


Tan  ^  =  -  =  .5206, 

2i        a 

Tan  27|°  =  .5206, 

=  .96"  =  C"  for  a  55°  sharp  thread. 


Then  the  depth  of  a  thread  of  £"  is  £  of  .96",  or  .12".  • 
For  a  U.S.  standard-shaped  thread  the  same  principle 

holds.  The  angle  is  the  same 

as  for  a  V-shaped  thread, 

namely  60°.    Since  -J  P  (see 

Fig.  93)  is  taken  off  the  top 

and  ^  filled  in  at  the  bottom, 

its  depth  is  only  f  of  that  of 

a  V-thread,  that  is,  C'  =  f  C.       FlG-  93«    U'S-  STANDARD  THREAD 

Hence,  by  remembering  C  for  a  screw  of  1"  pitch,  the  depth,  double 
depth,  and  root  diameter  of  any  V-shaped  or  U.S.  standard-shaped 
thread  may  be  easily  found. 


150 


SHOP  PROBLEMS  IN  MATHEMATICS 


PROBLEMS 

1.  A  screw  has  12  threads  per  inch.    What  is  its  pitch  ? 
What  is  its  lead  ? 

2.  A  lathe  lead  screw  has  2^  threads  per  inch.    What  is 
its  lead  ? 


3.  A  certain  pipe  thread  has  11^  threads  per  inch.  What 
is  its  pitch  ? 

4.  Find  the  pitch  for  the  following  number  of  threads 
per  inch  :  3,  3^,  4,  4^-,  57  5^-,  6,  6£. 

5.  Find  the  number  of  threads  per  inch  for  the  following 

leads  :  r,  t",  r,i",  F,  r,  i". 

6.  Denoting  the  number  of  threads  per  inch  by  TV,  the 
body  diameter  by  D,  and  the  pitch  by  P,  write  a  formula 
for  P  in  terms  of  N. 

7.  Solve  the  formula  in  Problem  6  for  N. 

8.  Find  the  value  of  C  for  a  screw  of  1"  pitch.    (Use  the 
geometrical  method.) 

9.  Make  a  table  for  both  V-  and  U.S.  standard  threads 
as  follows  : 


THREAD  PROPORTIONS, 

V-SHAPE 

Number 
of  threads 
per  inch 

Pitch 

Double 
depth 

THREAD  PROPORTIONS, 
U.S.  STANDARD 

Number 
of  threads 
per  inch 

Pitch 

Double 
depth 

THREAD  PROPORTIONS  151 

10.  The  depth  of  a  certain  thread  is  .025".    What  is  the 
root  diameter  of  a  screw  whose  outside  diameter  is  1^"  ? 

11.  If  the  double  depth  of  a  thread  is  .108",  to  what  size 
must  a  nut  be  bored  to  receive  a  screw  whose  outside 
diameter  is  J"  ? 

12.  Find  the  depth  of  thread  for  the  common  V-thread 

r»ifr»>if»Q      1    "       1    "       1    "       1    "       1    "       1    "       1        1"     1"     1"     1" 
pltCiieS  ^0    >  I*g    >  iV    >  T4    >   T3    )   T2    9    10)    8    >   ff    >    5    ?  ¥    • 

13.  Find  the  depth  of  thread  for  the  pitches  given  in 
Problem  12,  but  for  a  U.S.  standard-shaped  thread. 

14.  Ascertain  the  root  diameter  of  a  screw  1^"  in  outside 
diameter  and  with  7  threads  per  inch ;  with  10  threads ; 
12  threads  ;  40  threads. 

The  common  method  of  expressing  the  size  of  a  screw  is  ^"-20. 
It  means  that  the  screw  body  is  J"  in  diameter  and  has  20  threads  to 
the  inch. 

15.  Find  the  root  diameter  of  the  following  V-shaped 
threads  :  J»-20,  &»-!*,  f  "-16,  TV'-14,  J»-12,  $"-11,  f  "-10, 
|"-9,  l"-8. 

16.  Find  the  root  diameter  of  the  following  U.S.  standard 
threads :  J»-20,  TV'-18,  J"-16,  TV '~14,  J"-13,  T^"-12,  |"-11, 
£"-10,  |"-9,  l"-8,  lJ"-7,  lJ"-6.    Tabulate. 

17.  Allowing  30,000  Ib.  to  the  square  inch  for  the  ten- 
sile strength,  find  the  load  which  a  V-shaped  screw,  f  "-10, 
would  hold. 

18.  Find  the  tensile  strength  of  a  U.S.  standard  bolt 
f  "-11,  allowing  30,000  Ib.  per  square  inch. 

19.  What  per  cent  stronger  is  a  screw  which  is  threaded 
with  a  U.S.  standard-shaped  thread  than  one  with  a  V- 
shaped  thread  ? 

The  point  of  a  tool  for  cutting  U.S.  standard-shaped  threads  is 
flattened  an  amount  equal  to  one  eighth  of  the  pitch  of  the  screw  to 
be  cut. 


152 


SHOP  PROBLEMS  IN  MATHEMATICS 


A.  Worm  Thread 


20.  Find  the  width  of  the  point  of  a  tool  for  cutting  a  U.S. 
standard  thread  of  i"  pitch;  J"  pitch;  ^" pitch;  ^"  pitch. 

21.  Find  the  width  of  the  point  of  a  tool  for  cutting  a 
U.S.   standard-shaped  thread  (double)   of  10  threads  per 
inch ;  8  threads  per  inch ;  6  threads  per  inch. 

22.  Find  the  tap  drill  for  a  1"  U.S.  standard  screw. 

23.  Find  the  tap-drill  size 
for  a  14-20  tap.  The  size  of 
No.  14  wire  is  .242"  and  the 
thread  is  V-shape. 

24.  Find  the  tap  drill  for  a 
12-24  tap.  The  size  of  No.  12 
wire  is  .215"  and  the  thread 
is  U.S.  standard  shape. 

25.  It  is  desired  to  cut  20 
threads  to  the  inch  on  a  shaft 
1.375"   in    diameter.     What 
will  be  the  root  diameter  ? 

26.  If  an  inside  thread  of 
18  to  the  inch  is  to  be  cut  in 
a  nut  to  be  fitted  to  a  1.75" 
shaft,  what  is  the  smallest 
hole   that   can   be    bored    in 
the  nut? 

27.  Denoting  the  outside  diameter  of  a  screw  by  D,  the 
number  of  threads  by  JV,  and  the  root  diameter  by  dr,  make 
a  formula  for  dr  in  terms  of  D,  N,  and  a  constant  C. 

28.  A  U.S.  standard  bolt  2"  in  diameter  has  4|-  threads 
per  inch%    How  much  larger  should  a  bolt  with  V-shaped 
threads  be  in  order  to  be  as  strong  as  the  U.S.  standard  ? 

Square  threads  are  shaped  so  that  the  depth  of  tooth,  width  of 
tooth,  and  width  of  space  between  the  teeth  are  equal.  Hence  the 
depth  of  thread  is  equal  to  i  P  (see  Fig.  94). 


B.    Square  Thread 
FIG.  94 


THREAD  PROPORTIONS  153 

29.  A  square-threaded  screw  has  4J  threads  to  the  inch. 
How  wide  must  the  point  of  a  tool  be  made  to  cut  this 
thread  ? 

30.  A  shaft  3-J-"  in  diameter  has  a  square  thread  cut  upon 
it,  1^  threads  to  the  inch.    What  is  the  root  diameter  ? 

31.  A  nut  is  to  be  made  to  fit  a  square-threaded  screw 
of  JQ"  pitch  and  of  £"  diameter.    What  width  of  tool  should 
be  used  ?    What  diameter  should  the  hole  be  bored  ? 

32.  Compare  the  shearing  strength  of  the  threads  of  a 
U.S.  standard  screw  and  of  a  square-threaded  screw,  diam- 
eter of  the  shaft  being  1  j-",  with  6  threads  per  inch. 

For  the  proportions  of  a  worm  thread  see  Fig.  94. 
118.  Proportions  of  an  Acme  thread.   An  Acme  thread  is 
very  similar  to  a  worm  thread,  but  has  the  following  pro- 
portions : 

Let  Da  =  outside  diameter. 

N  =  number  of  threads  per  inch. 

Root  diameter  of  screw  =  Dn  —  I  --  h  .02  ). 

\N  ) 

Outside  diameter  of  tap  =  Da  +  .02. 
Depth  of  thread  =  -  -----  h  .01. 

Width  of  point  of  tool  for  screw  =  '~^  -  .0052. 
Width  of   screw  thread  at  top   =  : 


PROBLEMS 

1.  Denoting  the  outside  diameter  by  Da,  the  root  diam- 
eter by  dr,  and  the  pitch  by  P,  write  a  formula  for  the  root 
diameter,  drj  of  an  Acme  screw  in  terms  of  Da  and  P. 

2.  An  Acme  thread  is  to  be  cut  on  a  2-J-"  shaft.    If  there 
are  to  be  3  threads  per  inch,  what  will  be  the  root  diameter  ? 


154  SHOP  PROBLEMS  IN  MATHEMATICS 

3.  A  nut  is  to  be  threaded  to  fit  a  screw  If"  in  diameter, 
with  4|  threads  per  inch.    How  large  a  hole  must  be  bored 
in  the  nut,  allowing  .050"  for  clearance  ? 

4.  Find  the  width  of  the  point  of  an  Acme-thread  tool 
for  5  threads  per  inch  ;  4  threads  ;  3  threads  ;  1^  threads. 

119.  Double  threads,  triple  threads,  etc,'  Oftentimes  the 
depth  of  a  thread  upon  a  shaft  is  great  enough  to  materially 
reduce  its  strength.  When  an  extreme  lead  is  required  for 
a  screw  and  the  shaft  must  not  be  materially  weakened, 
a  double,  or  triple,  or  even  quadruple  thread  may  be  cut. 

In  a  double  thread  the 
3.2JA2J.          .    •_         ,    ,      . 

pitch  and  depth  are  only 

one  half  of  that  of  a  single 
thread  of  the  same  lead. 
Similarly,  for  a   triple 
thread  the  pitch  and  depth 
are  only  one  third  of  that 
FIG.  96.  SINGLE  AND  TRIPLE  THREADS    of  a  single  thread  of  the 
or  THE  SAME  LEAD  same  iead 

Fig.  95  shows  a  single-  and  a  triple-threaded  screw  of  the  same 
lead.  Note  the  excessive  depth  in  the  single  thread.  Also  note  the 
three  endings  of  the  triple  thread. 

PROBLEMS 

1.  A  double-threaded    screw  has  12  threads  per   inch. 
What  is  its  pitch  ?    What  is  its  lead  ? 

2.  Find  the  double  depth  of  the  screw  in  Problem  1. 

3.  A   triple-threaded    screw   has    15    threads   per   inch. 
What  is  its  pitch  ?    What  is  its  lead  ? 

4.  A   double-threaded   screw   has   9   threads   per    inch. 
What  is  the  lead  and  the  equivalent  single  thread  ? 

5.  A  double-threaded  screw  of  6  threads  per  inch  and  'a 
triple-threaded  screw  of  9  threads  per  inch  have  the  same 


GEARING  FOR  SCREW  CUTTING  155 

lead.    What  per  cent  stronger  would  a  1^"  shaft  be  with  a 
triple  than  with  a  double  thread  cut  upon  it  ? 

Unless  otherwise  stated  the  U.S.  standard-shaped  threads  should 
be  taken. 

6.  Compare  the  strength  of  a  screw  when  cut  with  a 
single,  double,  or  triple  thread  of  £"  lead  on  a  1"  shaft, 
allowing  a  tensile  strength  of  60,000  Ib.  per  square  inch. 

7.  A  double-  and  a  triple-threaded  screw  have  £"  lead. 
What  number  of  threads  do  they  have  per  inch?    What 
are  their  depths  of  threads  ? 

8.  A  nut  is   to  be  bored  out  preparatory  to  cutting  a 
triple  thread  of  J"  lead  for  a  3"  shaft.    What  size  should 
the  hole  be  made  ? 

9.  Find  the  root  diameter  of  a  quadruple-threaded  screw 
of  f "  lead  and  1£"  outside  diameter. 

10.  Find  the  lead  for  a  double-threaded  screw  which  will 
have  the  same  depth  as :  single  threads  of  yL"  pitch,  £" 
pitch,  I"  pitch,  T-y  pitch  respectively;  triple  threads  of  y1^" 
pitch,  |"  pitch,  J"  pitch  respectively. 

For  the  method  of  setting  tools  for  multiple  threads  see  the  fol- 
lowing sections. 

GEARING  FOR  SCREW  CUTTING 

120.  Simple  gearing.  A  simple  geared  lathe  is  illus- 
trated in  Fig.  96.  The  change  gears  are  the  gears  on  S 
and  L.  Gear  7  is  used  only  to  connect  S  and  L,  and  does 
not  appear  in  the  calculations.  The  reverse  gears,  as  shown 
in  both  Figs.  96  and  97  in  their  two  different  positions, 
are  simply  for  the  purpose  of  changing  the  direction  of 
rotation  of  gear  L,  and  hence  of  the  lead  screw.  Thus  by 
reversing  these  gears  a  right-  or  left-hand  thread  may  be  cut. 

The  first  thing  to  do  in  calculating  gearing  is  to  see  that  spindle  S' 
makes  the  same  number  of  revolutions  as  S.  If  they  do  not  revolve 


156 


SHOP  PROBLEMS  IN  MATHEMATICS 


the  same,  then  equal  gears  should  be  placed  on  S  and  L  and  the  lathe 
tested  to  see  what  screw  would  be  cut.  This  screw  should  then  be 
used  as  the  calculating  lead  screw  instead  of  the  lead  screw  on  the 
machine.  (By  "screw"  we  mean  the  number  of  threads  per  inch.) 

121.  The  rules  for  calculating  the  gears  for  thread  cutting. 

RULE  I.    The  number  of  threads  per  inch  on  the  screw  to 

be  cut  is  to  the  number  of  threads  per  inch  on  the  lead  screw 


Reverse 
Gears 


Intermediate 

Gear  on  Screw 


FIG.  96.    END  VIEW  OF  LATHE  HEADSTOCK 

as  the  number  of  teeth  in  the  gear  on  the  lead  screw  is  to  the 
number  of  teeth  in  the  gear  on  spindle  or  stud; 
Screw  to  be  cut  _  Gear  on  lead  screw 

Lead  screw  Gear  on  spindle 

For  example,  find  the  gears  for  cutting  a  thread  of  -J" 
pitch  on  a  lathe  whose  lead  screw  has  5  threads  per  inch. 


GEARING  FOR  SCREW  CUTTING 


157 


-J"  pitch  is  equivalent  to  8  threads  per  inch.    Our  ratio  is 
8      Gear  on  lead  screw 
5         Gear  on  spindle 

But  since  there  are  no  gears  of  5  or  8  teeth  we  are 
obliged  to  multiply  the  terms  of  the  ratio  by  some  num- 
ber, as  4,  5,  G,  or  7. 


Compound 
Gearing 


Then 


FIG.  1>7.    END  VIEW  OF  LATHE  HEADSTOCK 


8      6      48      Gear  on  lead  screw 


5X6      30 


Gear  on  spindle 


If  gears  of  48  or  30  teeth  are  not  available,  then  take  some  other 
number  than  6  as  a  multiplier. 

RULE  II.  If  the  screw  to  be  cut  is  a  larger  number  than 
the  lead  screw,  then  the  smaller  gear  goes  on  the  spindle,  or 
vice  versa. 


158  SHOP  PROBLEMS  IN  MATHEMATICS 

122.  Compound  gearing.  When  by  simple  gearing  the 
number  of  teeth  in  the  gears  run  up  to  over  120,  com- 
pounding is  resorted  to.  That  is,  an  additional  gear,  C 
(see  Fig.  97),  is  put  upon  a  sleeve  along  with  D,  or  a 
small  bracket,  C'  (see  Fig.  96),  holding  two  gears,  may 
be  swung  into  mesh  between  S  and  /.  The  change  gears 
in  Fig.  97  are  A,  C,  D,  and  /,. 

The  rules  for  calculating  the  compound  gears  for  thread 
cutting  are  as  follows  : 

T-i-   •  7       7  /.  Screw  to  be  cut   . 

RULE  III.   Divide  tlie  ratio  of    —  -  -  into  two 

Lead  screw 

ratios,  the  terms  of  which  may  be  increased  Inj  a  multiplier 
to  such  numbers  as  will  represent  available  gears. 

RULE  IV.  In  compounding,  if  the  denominator  of  one  ratio 
is  taken  as  a  driving  gear,  then  the  denominator  of  the  other 
ratio  will  be  a  driving  gear  and  the  numerators  will  be 
driven  gears,  or  vice  versa. 

For  example,  find,  the  gearing  necessary  to  cut  22  threads 
per  inch  on  a  lathe  with  a  lead  screw  of  5  threads  per  inch. 

Our  ratio  is  ^2  =  y  X  ?. 

Bracket  C"  (see  Fig.  96)  has  two  gears  fixed  upon  it,  generally  of 
60  and  30  teeth  ;  therefore  one  of  our  ratios  should  be  multiplied  by 
such  a  number  as  to  give  60  and  30. 


The  required  gears  then  are  §§  and 
.  It  is  not  so  easy  to  determine  their  relative  positions  on 
the  lead  screw  or  spindle  as  in  simple  gearing,  but  it  may 
be  arrived  at  according  to  Rules  II  and  IV. 

Since  22  is  larger  than  5,  the  gears  30  and  30  will  be  the 
drivers  and  gears  66  and  60  the  driven  gears. 

In  Fig.  97  gears  A  and  D  are  drivers  and  C  and  L  are  driven  gears. 
If  the  ratios  obtained  at  first  do  not  represent  available  gears,  new 
ratios  must  be  taken. 


GEARING  FOR  SCREW  CUTTING  159 

123.  Metric  threads.  A  similar  procedure  may  be  fol- 
lowed in  rinding  the  gears  necessary  to  cut  a  metric  thread 
without  a  metric  lead  screw. 

For  example,  find  the  gears  necessary  to  cut  a  thread 
of  1.5  mm.  pitch  on  a  lathe  with  a  lead  screw  of  5  threads 
per  inch. 

This  pitch  must  first  be  reduced  to  threads  per  inch. 
1"  =  25.4  mm.  ; 

25.4 
hence  -T-TT-  =  16}  |  threads  per  inch. 

16}*-      254 

Our  ratio  is  '     —  -=^-' 

o  75 

Since  in  these  calculations  the  figure  254  will  always  appear  no 
matter  what  the  pitch,  and  since  its  factors  are  127  and  2,  it  follows 
that  127  should  always  be  selected  as  the  term  of  one  ratio. 


Hence  our  gears  are  -^V  and  f  g. 

Then  25  and  60  are  the  drivers  and  127  and  40  are  the 
driven  gears. 

/PROBLEMS 
Find  the  gears  necessary  to  cut  a  screw  of  10  threads 
per  inch  on  a  lathe  with  a  lead  screw  of  5  threads  per  inch. 
Select  from  the  following  gears  :  25,  30,  35,  40,  45,  50,  55, 
60,  65,  69,  70,  80,  90,  100,  120. 

2.  The  lead  screw  of  a  lathe  has  6  threads  per  inch. 
What  gears  will  be  necessary  to  cut  8  threads  ? 

Do  not  use  a  gear  of  less  than  24  teeth. 

3.  On  the  lathe  of  Problem  1  what  is  the  largest  num-/ 
ber  of  threads  that  can  be  cut  by  simple  gearing  ?    The 
gear  of  120  teeth  must  be  used  as  an  intermediate. 

4.  The  standard  thread  for  f  "  pipe  is  ll.£  per  inch.   What 
gears  are  necessary  to  cut  this  thread  with  a  lead  screw  of 


160  SHOP  PROBLEMS  IN  MATHEMATICS 

6  threads  per  inch  ?    5  threads  per  inch  ?    4  threads  per 
inch  ?    Use  69  as  the  gear  on.  the  lead  screw. 

5.  A  lathe  with  a  lead  screw  of  5  threads  per  inch  is 
supplied  with  gears  as  given  in  Problem  1.    Can  a  screw 
of  2  threads  per  inch  be  cut  on  this  lathe  ? 

6.  With  the  gears  given  in  Problem  1,.  find  the  correct 
gears  for  cutting  24  threads  by  compounding. 

7.  The  calculating  lead  screw  of  a  lathe  has  16  threads 
per  inch,  and  the  lathe  is   equipped  Avith  the  following 
gears :  32,  32,  36,  40,  44,  46,  48,  52,  56,  64,  72,  80,  96, 
120.    Select  the  sets  of  gears  necessary  to  cut  4  threads ; 
5  threads  ;  6  threads. 

8.  Using   the  lathe  as   described  in  Problem   7,  select 
proper  gears  for  cutting  8  threads  ;  20  threads  ;  36  threads. 

9.  The  calculating  lead  screw  of  a  lathe  is  8.    Select  gears 
from  the  following  suitable  to  cut  11^  threads  per  inch : 
24,  24,  28,  32,  32,  36,  40,  44,  48,  52,  56,  64,  69,  72,  80,  96. 

10.  From  the  gears  in  Problem  9  select  the  sets  of  gears 
necessary  to  cut  the  following  threads  per  inch:   1,  14-,  2, 
3,  4,  5,  using  as  the  compound  the  gears  ||. 

11.  What   is   the   smallest  thread  that  can  be  cut  by 
simple  gearing,  using  the  lathe  and  gears  of  Problem  9  ? 

12.  The  ratio  of  the  K.  P.  M.  of  the  lathe  spindle  to  those 
of  the  stud  (see  Fig.  96)  is  4  to  3.    Find  the  calculating 
lead  screw  if  the  real  lead  screw  has  6  threads  per  inch. 

Place  equal  gears  on  stud  and  lead  screw. 

13.  Find  the  gears  necessary  to  cut  a  metric  screw  of 
4  mm.  pitch  on  a  lathe  with  a  lead  screw  of  £"  pitch. 

I"  =  25.4  mm. 

14.  Find  the  number  of  threads  per  inch  which  is  equal 
to  1  mm.  pitch ;    1.5  mm.  pitch ;    2  mm.  pitch ;   2.5  mm. 
pitch  ;  3  mm.  pitch. 


INDEXING 


161 


c— 


15.  Find  the  gears  necessary  to  cut  a  metric  screw  of 
2  nun.  pitch  on  a  lathe  with  a  lead  screw  of  5  threads  per 
inch.    Do  not  use  a  gear  of  less  than  25  teeth. 

16.  How  many  metric  threads  could  be  cut  by  the  fol- 
lowing gears  :  127,  25,  40,  60  ?    What  are  their  pitches  ? 

17.  Find  the  gears  necessary  to  cut  a  metric  screw  of  3.5 
mm.  pitch  on  a  lathe  with  a  lead  screw  of  6  threads  per  inch. 

124.  Indexing.    It  is  frequently  necessary  to  space  equal 
distances  upon  a  surface,  as  holes  in  a  circle  or  teeth  upon 
a  gear,  or  to  mill  an  hexagonal 

or  square  nut.  Locating  the  drill 
or  cutter  for  these  purposes  is 
properly  called  indexing.  This 
work  is  most  frequently  done 
upon  a  milling  machine,  although 
the  principle  may  be  applied  to 
any  machine. 

125.  Indexing  by  use  of  a  gear. 
The  teeth  of  a  gear  may  furnish 
the    means    of    indexing.    This 
gear  may  be  put  on  the  cross- 
feed  screw  of  a  shaper  or  planer,  and  thus  enable  one  to  cut 
a  rack,  etc.,  upon  these  machines.    A  disk  with  a  circle  of 
equally  spaced  notches  may  be  placed  upon  the  back  end 
of  a  lathe  spindle  and  indexing  done  upon  the  lathe. 

126.  Index  head.   This  head  consists  essentially  of  a  base 
B  (see  Fig.  98),  a  spindle  (7,  revolved  by  a  worm  wheel 
W  and  worm.     Arm  A    and  the    index    plate  I  are   con- 
nected to  the  worm  shaft.    The  index  plate  generally  has 
a  number  of  circles  of  holes  called  index  circles,  each  circle 
having  a  different  number  of  holes,  as  31,  33,  37,  39,  etc. 

It  is  by  means  of  these  index  circles  that  we  are  enabled 
to  get  an  exact  fraction  of  a  revolution,  as  J§,  this  being 


FIG.  98.    INDEX  HEAD 


162  SHOP  PROBLEMS  IN  MATHEMATICS 

13  holes  in  a  33  circle ;  or  f  f ,  being  25  holes  in  a  27  circle. 
Sometimes  an  index  plate  is  attached  directly  to  spindle  C ; 
then  each  circle  will  index  a  few  numbers  directly.  For 
instance,  a  circle  of  30  holes  would  index  2,  3,  5,  6, 10, 15, 30. 

127.  Method.  It  takes  forty  revolutions  of  arm  A  to  turn 
spindle  C  around  once.  Hence  it  can  be  seen  that  a  rela- 
tively large  movement  of  arm  A  produces  only  a  small 
rotation  (?L)  of  spindle  C.  For  instance,  if  we  rotate  A 
TL  of  one  revolution,  C  will  revolve  ^  of  one  revolution; 
or  if  we  rotate  A  13^  times,  then  C  will  revolve  ^  of  13£, 
or  ^  of  one  revolution,  so  that  by  moving  arm  A  this  amount 
3  times  in  succession  we  are  able  to  obtain  3  equal  spaces 
in  one  revolution. 

RULE  I.  Divide  40  by  the  number  to  be  indexed  and  the 
result  will  be  the  turns  required  for  arm  A. 

RULE  II.  The  denominator  of  the  fraction  determines  the 
index  circle  to  be  used\ 

RULE  III.  If  the  fraction  is  a  small  one,  its  terms  must 
be  multiplied  by  some  number  which  will  give  an  available 
circle. 

For  example,  find  the  number  of  turns  necessary  to  cut 
22  teeth  in  a  gear. 

*tf  —  lit  =  ly9!  —  number  of  turns. 

T9T  X  I  =  If.     T\  X  g  -  §J. 

There  are  two  index  circles  which  may  be  used  in  this 
ca^e,  namely  22  or  33. 

PROBLEMS 


V 


1.  An  index    plate  on  spindle  C  (see  Fig.  98)  has  24 
holes.    What  numbers  can  be  indexed  by  it  ? 

2.  A  certain  milling  machine  is  supplied  with  three  in- 
dex plates  containing  the  following  circles :  15,  16,  17,  18, 
19,  20,  21,  23,  27,  29,  31,  33,  37,  39,  41,  43,  47,  49.    Find 


INDEXING  163 

the  index  circle  and  the  number  of  turns  necessary  to  index 
these  numbers :  5,  7,  9,  10,  11,  13,  28,  29,  88,  98,  100,  260. 

3.  A  reamer  is  to  have  8  teeth,  unequally  spaced  one  to 
the  other.    Present  a  scheme  by  which  this  can  be  done  on 
the  index  head. 

4.  Find  the  different  numbers  up  to  100  which  can  be 
indexed  by  using  the  33  circle. 

5.  What  numbers  from  1  to  300  can  be  indexed  in  com- 
mon by  the  27  and  33  circles  ? 

6.  Find  the  three  circles  which  will  index  in   common 
the  following  numbers :  3,  15,  30,  59,  61. 

A  straight  piece  of  iron  or  steel  with  gear  teeth  cut  in  it  is  called 
a  rack. 

7.  Several  teeth  in  a  rack  of  .3142"  circular  pitch  must 
be  recut  upon  a  shaper  whose  cross-feed  screw  has  4  threads 
per  inch.    A  gear  of  86  teeth  is  mounted  on  the  end  of  the 
screw  for  indexing.    How  many  turns  will  be  necessary  to 
move  the  tool  from  one  tooth  to  the  next  ? 

A  slight  approximation  must  be  made. 

8.  What  error  in  thousandths  of  an  inch  is  made  in  the 
spacing  of  each  tooth  by  the  approximation  made  in  Prob- 
lem 7  ? 

9.  Would   a   gear  of   100   teeth  give  a  more   accurate 
spacing  for  the  rack  in  Problem  7  than  a  gear  of  86  teeth  ? 
Why? 

10.  What  gear  among  the  following  would  give  more 
accurate  spacing  for  the  rack  in  Problem  7  than  the  gear  of 
86  teeth :  25,  30,  35,  40  ?   What  error  in  thousandths  would 
this  gear  make  in  the  spacing  of  each  tooth  ? 

11.  The  following  gears  are  included  with  the  equipment 
of  a  lathe :  25,  30,  35,  40,  45,  50,  55,  60,  65,  69,  70,  80,  90. 
Find  a  gear  which  may  be  used  for  spacing  the  teeth  of  a 


164  SHOP  PROBLEMS  IN  MATHEMATICS 

rack  of  .7854"  circular  pitch  upon  the  shaper  of  Problem  7  ; 
.5236"  circular  pitch  ;  f  "  circular  pitch  ;  3  diametral  pitch  ; 
16  diametral  pitch. 

The  circular  pitch  is  equal  to  TT  (V)  divided  by  the  diametral  pitch. 

12.  A  punch  shaped  as  shown  in  Fig.  99  is  to  have  the 
three  surfaces,  A,  B}  C,  milled  with  a  plain"  mill.  Find  the 
index  circla  which  may  be  used  and  the  number  of  turns 
necessary  to  bring  B  into  a  horizontal  position  after  milling 
A.  Also  find  the  turns  necessary  to  bring  C  into  the  hori- 
zontal position  after  milling  B. 

128.  Setting  the  tool  in  cutting  double  threads,  etc.    In 
double,  triple,  or  quadruple  threads,  after  cutting  the  first 
thread  the  tool  must  be  put  into  position 
for  the  next  thread.    This  can  be  done 


,  60' 

in  several  ways. 


(1)  By  changing  the  tail  of  the  dog  in      FlG  99    END  V,EW 
the  face  plate.   The  face  plate  must  have  OP  PUNCH 
slots  accurately  milled  for  this  method. 

(2)  By  careful  measurement  and  resetting  of  the  tool. 

(3)  By  withdrawing  gear  7  (see  Fig.  96)  and  revolving 
gear  S  £,  ^,  or  ^  of  a  complete  revolution  and  again  placing 
/  in  mesh  with  S.    This  moves  the  spindle  but  not  the 
carriage,  and  can  only  be  used  when  the  gear  S  is  divisible 
by  2  for  double  threads  or  3  for  triple  threads,  etc. 

(4)  By  calculating  how  many  turns  of  gear  L  would  be 
necessary  to  move  the  carriage  the  required  distance  and 
by  withdrawing  /  while  L  is  being  turned  this  amount. 

PROBLEMS 

1.  Find  the  gears  necessary  to  cut  a  double-threaded 
screw  of  £"  lead  on  a  lathe  with  a  lead  screw  of  5  threads 
per  inch.  Use  5  as  a  multiplier. 


INDEXING  165 

2.  Can  the  tool  be  set  for  the  second  thread  by  the  third 
method  described  in  section  128  ? 

3.  A  triple-threaded  screw  of  6  threads  per  inch  is  to  be 
cut  011  a  lathe  with  a  lead  screw  of  4  threads  per  inch. 
Find  the  gears  required  when  the  smallest  gear  has  30  teeth. 

4.  A  lathe  is  set  up  with  the  70-tooth  gear  on  the  screw 
and  the  28-tooth  gear  on  the  stud.    What  thread  would  be 
cut  if  the  lead  screw  were  of  £"  lead  ? 

5.  A  double-threaded  screw  of  i"  pitch  is  to  be  cut  on  a 
lathe  having  a  lead  screw  of  £"  pitch.    Select  the  proper 
gears  from  the  following :  25,  30,  35,  40,  45,  50,  60,  75,  90. 

6.  Can  the  tool  be  set  for  the  second  thread  by  the  third 
method  in  section  128  ?    Why  ? 

7.  Find  the  number  of  turns  of  gear  L  (see  Fig.  96)  neces- 
sary to  set  the  tool  for  the  second  thread  in  Problem  1. 

8.  A  triple  thread  of  £"  pitch  is  to  be  cut  on  a  lathe 
having  a  lead  screw  of  £"  pitch.    Select  the  proper  gears 
from  those  given  in  Problem  5. 

9.  Find  the  number  of  turns   of  the  gear  on  the  lead 
screw  in  order  to  set  the  tool  to  the  correct  position  for 
the  second  thread  (see  sect.  128). 

10.  How  much  must  the  tool  be  advanced  in  setting  it 
for  the   second  thread   on   a  double-threaded  screw  of  3 
threads  per  inch  ?  4  threads  ?  6  ?  7  ?  8  ? 

11.  How  much  must  the  tool  be  advanced  in  setting  it 
for  the  second  thread  on  a  triple-threaded  screw  of  3  threads 
per  inch  ?  4  threads  ?  5  ?  6  ?  7  ? 


CHAPTER   XI 

GEAR  PROPORTIONS  AND  SPIRALS 

CALCULATIONS  OF  GEAR  TEETH 

129.  Speed  ratio.    The  estimating  of  all  gearing  is  based 
upon  such  theoretical  cylinders  (for  spur  gears)  or  cones  (for 
bevel  gears)  as  will  produce  the  desired  speed  ratio  when 
rolled  upon  each  other,  no  slip  being  considered. 

130.  Pitch   circles   and   pitch   diameters.    The  circumfer- 
ences of  these  cylinders  are  called  pitch  circles  and  their 
diameters   pitch  diameters.     In   gearing,  the   pitch   circle 
is  located  practically  halfway  between  the  top  and  bottom 
of  the  tooth. 

131.  Spur  gears.    Spur  gears  are  gears  cut  on  a  cylinder 
with  teeth  parallel  to  the  axis. 

132.  Spiral  gears.    Spiral  gears  have  their  teeth  cut  at  an 
angle  to  the  axis  of  the  cylinder. 

133.  Bevel  gears.    Bevel  gears  are  gears  cut  upon  a  con- 
ical surface. 

134.  Diametral  pitch.    The  number  of  teeth  in  a  gear  per 
inch  of  its  pitch  diameter  is  called  the  diametral  pitch. 

135.  Circular  pitol£    The  distance  from  the  center  of  one 
tooth  to  the  center  of  the  next  measured  on  the  pitch  circle 

is  called  the  circular  pitch,  and  is  equal  to  -J— 

Jd 

136.  Addendum.    The  distance  from  the  pitch  circle  to  the 
outside  of  the  tooth  is  called  the  addendum.    It  is  equal  to 

1  divided  by  the  diametral  pitch,  or  —  (see  sect.  145). 

l<i 
166 


GEAR  PROPORTIONS  AND  SPIRALS  167 

137.  Dedendum.    The  distance  from  the  pitch  circle  to 
the  bottom  of  the  tooth  is  called  the  dedendum,  and  is  also 

equal  to  — • 

*d 

138.  Outside  diameter.    The  outside  diameter  is  equal  to 
the  pitch  diameter  plus  twice  the  addendum. 

139.  Number  of  teeth.    The  number  of  teeth  by  definition 
is   equal  to  the  diametral  pitch  multiplied  by  the  pitch 
diameter. 

140.  Depth  of  the  tooth.    The  depth  of  the  tooth,  not  in- 

2 
eluding  clearance,  is  2  s,  or  —  • 

*d 

141.  Clearance.    The  clearance  (/)  allowed  at  the  bottom 
of  the  tooth  is  equal  to  TL  of  the  thickness  (£)  of  the  tooth 

at  the  pitch  circle. 

2 

142.  Whole  depth  of  the  tooth  is  equal  to  2  s  +/,  or  -  +/. 

A/ 

143.  The  thickness  of  a  tooth  on  the  pitch  circle  is  equal 
to  the  circumference  of  the  pitch  circle  divided  by  twice  the 
number  of  teeth,  or  the  circular  pitch  divided  by  2. 

144.  Center  distances.    The  distance  between  the  centers 
of  two  gears  is  equal  to  one  half  the  sum  of  their  pitch 
diameters.    Or,  if  the  pitch  diameters  are  unknown,  the  dis- 
tance may  be  obtained  from  the  number  of  teeth  N  and  Nl 
in  each  gear  and  the  diametral  pitch,  both  of  which  are  gen- 
erally known. 

In  the  following  notes  and  problems  tbe  different  terms  used  are 
indicated  in  the  formulas  by  letters  as  shown  in  section  145.  In  order 
that  the  pupil  may  solve  all  the  problems  given  below,  he  should  tabu- 
late the  formulas  for  the  different  tooth  parts  as  he  works  them  out. 

145.  Notation. 

Diametral  pitch  =  Pd. 

Circular  pitch  =  Pc. 

Pitch  (circle)  diameter  =  D. 


168  SHOP  PROBLEMS  IN  MATHEMATICS 

Circumference  of  pitch  circle  =  TrD. 

Number  of  teeth  in  gear  =  N. 

Addendum  =  s. 

Dedendum  =  s. 

Distance  between  centers  of  2  gears  =  C. 

Thickness  of  tooth  at  pitch  circle  =  t. 

Clearance  at  bottom  of  tooth  =f. 

Outside  diameter  =  OD. 

Working  depth  =  W. 

Whole  depth  =  W'. 

146.  Derivation  of  formulas. 

EXAMPLE.    Derive  the  formula  for  N  when  the  outside 
diameter  and  the  diametral  pitch  are  given. 

From  section  139  we  find  that  N  =  P(,D(V),  and  from  sec- 
2 


tion  138  that  OD  =  D  +  —  (2).   Find  N  in  terms  of  OD  and 

* 

Pd.   Eliminate  D  from  equation  (1)  by  substituting  its  value 

2 

as  found  in  equation  (2),  that  is,  D  =  OD  --  -  • 

*ti 
Then  equation  (1)  is 

N  =  (OD  -    \pd  =  OD  x  Pd  -  2. 


EXAMPLE.    Derive  a  formula  for  t  in  terms  of  Pd.    Method 
of  solution.    By  definition 


DX  7T         7T         D 

hence  **  __=_.X-- 

But  7?,  =  ~ 

TT      1       3.1416      1.57 
/Hence  t  =  -  x  -=  ^-- 

Tabulate  these  formulas  when  solved. 


GEAR  PROPORTIONS  AND  SPIRALS  169 

PROBLEMS 

1.  Derive  a  formula  for  Pc  in  terms  of  N  and  D. 

2.  Find  the  Pc  of  a  gear  of  60  teeth  and  6"  pitch  diameter. 

3.  Derive  a  formula  for  Pd  in  terms  of  N  and  D. 

4.  A  gear  has  60  teeth  and  a  pitch  diameter  of  6".  What 
is  its  Pd  ? 

5.  Derive  Pd  in  terms  of  TT  and  Pc. 

6.  Derive  Pc  in  terms  of  TT  and  Pd. 

7.  Derive  N  in  terms  of  Pd  and  D. 

8.  If  a  gear  is  to  be  10  pitch  and  is  to  have  a  pitch  diam- 
eter of  2i",  how  many  teeth  will  it  have  ? 

9.  Derive  a  formula  for  D  in  terms  of  N  and  Pd. 

10.  Derive  a  formula  for  C  in  terms  of  N,  N',  and  7J. 


Where  the  word  "pitch"  is  used  alone,  the  diametral  pitch  is 
always  understood. 

12.  A  gear  blank  is  4f  "  in  outside  diameter  and  is  to  be  cut 
with  a  12-pitch  gear  cutter.    How  many  teeth  must  it  have  ? 

13.  A  gear  has  25  teeth  and  its  diametral  pitch  is  16. 
What  is  its  pitch  diameter  ?    What  is  its  outside  diameter  ? 

14.  Given  a  gear  of  3.222"  pitch  diameter  and  a  pitch  of 
9.   What  is  the  outside  diameter  ? 

15.  Given  a  gear  of  1.889"  pitch  diameter  and  36  pitch, 
find  the  outside  diameter.    Find  the  number  of  teeth. 

16.  Two  gears  have  70  and  60  teeth  respectively,  and  are  of 
10  diametral  pitch.   Find  the  distance  between  their  centers. 

17.  A  gear  has  a  pitch  diameter  of  '3T55'r  and  41  teeth. 
Find  the  outside  diameter. 

18.  Given  44  teeth  in  a  gear  of  10  diametral  pitch.   Find 
the  pitch  diameter. 


170  SHOP  PROBLEMS  IN  MATHEMATICS 

19.  What  is  the  outside  diameter  of  the  gear  mentioned 
in  Problem  18  ? 

20.  What  is  the  diameter  at  the  bottom  of  the  teeth, 

if  the  depth  of  tooth  below  the  pitch  circle  is  —  and  the 

1  * 

clearance  is  ^— -  ? 

21.  If  the  size  of  the  hole  for  the  shaft  in  the  gear  of 
Problem  18  is  1^-",  what  is  the  area  of  the  metal  left  to  resist 
the  breaking  of  the  gear?   The  width  of  the  gear  face  is  1". 

22.  Would  the  gear  break  at  the  tooth  or  in  the  body,  if 
overloaded  ? 

Allow  a  thickness  of  the  tooth  equal  to  %  the  circular  pitch. 

23.  How  large  would  the  hole  in  the  gear  have  to  be,  in 
order  that  the  gear  might  break  in  the  body  instead  of  in 
the  tooth  ? 

24.  A  gear  of  .3142"  circular  pitch  has  37  teeth.    Find  the 
pitch  diameter. 

There  are  two  methods  of  procedure. 

25.  A  gear  has  28  teeth  and  a  pitch  diameter  of  8".   What 
is  the  diametral  pitch  ? 

26.  Could  we  have  a  gear  of  the  same  pitch  as  in  Prob- 
lem 25  with  29  teeth  ? 

27.  A  broken  gear  has  24  teeth  and  approximately  If" 
outside  diameter.    What  is  the  diametral  pitch  ? 

28.  Two  gears  mesh  together.    One  has  28  teeth  and  an 
outside  diameter  of  2^";   the  other  has  68  teeth  and  an 
outside  diameter  of  5|".    Find  the  pitch  diameter  of  each. 
Find  the  center  distance. 

29.  A  gear  has  been  set  up  in  the  milling  machine  ready 
for  cutting  the  teeth.    The  pitch  is  to  be  14  and  the  pitch 
diameter  is  4".    What  is  the  depth  of  the  cut  ? 


GEAR  PROPORTIONS  AND  SPIRALS  171 

30.  A  gear  is  14  pitch  and  has  an  outside  diameter  of 
4j".    Find  the  depth  of  cut  for  the  teeth. 

31.  A  gear  has  75  teeth  and  is  18  pitch.    Find  the  work- 
ing depth  of  the  tooth. 

32.  The  circular  pitch  of  a  rack  is  .785".   Find  the  whole 
depth  of  the  tooth. 

33.  The  circular  pitch  of  a  rack  is  T5E  ".    Find  the  working 
depth  of  the  tooth.   What  movement  of  the  milling-machine 
table  will  be  necessary  to  space  these  teeth  ? 

34.  A  rack  is  to  be  cut  on  a  milling  machine  to  fit  a 
12-pitch  gear.    Find  the  depth  of  the  cut  and  the  longi- 
tudinal movement  of  the  table  for  spacing  the  teeth. 

35.  The  teeth  of  a  gear  8£"  outside  diameter  are  found  to 
be  8  diametral  pitch.    What  is  the  pitch  diameter  ?    What 
is  the  distance  between  centers  of  this  gear  and  one  of  4^" 
outside  diameter  ? 

36.  Make  a  formula  for  finding  the  pitch  diameter  Pd  when 
the  number  of  teeth  N  and  the  addendum  s  are  given. 

37.  Two  gears  are  placed  15^"  between  centers.   One  has 
9  teeth  and  the  other  66.    What  is  their  pitch  diameter  ? 

38.  In  turning  up  a  gear  blank  to  replace  one  that  is 
lost  we  find  from  the  gear  with  which  it  meshes  that  the 
diametral  pitch  is  7,  and  from  the  center  distances  that  the 
pitch  diameter  must  be  7".    What  size  must  the  gear  blank 
be  made  ?    What  will  be  the  number  of  teeth  ? 

39.  Given  one  gear  of  36  teeth  and  8  diametral  pitch,  and 
the  distance  9^"  between  centers.   Find  the  number  of  teeth 
and  the  outside  diameter  of  the  other  gear. 

40.  In  designing  a  set  of  gears  it  is  necessary  that  they 
shall  be  10"  between  centers  and  shall  have  a  speed  ratio 
of  1  to  3.    How  many  different  pitches  could  be  selected 
from  the  numbers  1  to  20  diametral  pitch  ?    Why  ? 


172 


SHOP  PROBLEMS  IN  MATHEMATICS 


41.  Two   gears    have   a   pitch   diameter  of   3.555"  and 
2.777".    AVhat  is  their  velocity  ratio  and  the  distance  be- 
tween their  centers  ? 

42.  Given  a  velocity  ratio  of  7  to  5  and  the  distance  6" 
between  centers.   Select  a  suitable  diametral  pitch  and  find 
the  remaining  data  necessary  to  construct  these  gears  in 
the  shop. 

It  is  necessary  for  the  draughtsman  in  designing  gears  to  know 
their  various  proportions  in  order  to  ascertain  their  strength.  Tables 
may  often  be  found  supplying  this  information,  but  the  student 
should  know  how  to  find  it.  It  is  recommended  that  a  table  be  made 
from  solutions  obtained  in  the  following  problems. 

43.  Find  the  circular  pitch  when  the  diametral  pitch  is 
i,  f,  1,  2,  3,  4,  5,  6,  7,  8,  9,  10,  12,  14,  16,  18,  20,  24,  30. 

44.  Find  the  diametral  pitch  when  the  circular  pitch  is 

9"      17"      13"      15"      11    It     13"     11"     11"      1"      7"     3"      5"      1"      7    " 

^    >    X¥    >    *--£    >    J-ff    J    ^2?    >    ^    )    -1?    >    -^    >    -1    J    *    >    I    J    B    5    2       T%    > 


2       157 

45.  Find  the  whole  depth  —  +  '——  for  each  of  the  above 

"d  *d 

diametral  and  circular  pitches. 

46.  Find  the  thickness  of  the  tooth  on  the  pitch  line  for 
the  above  pitches. 

Tabulate  the  results  obtained   in   Problems  43-46,  as 
follows  : 

SAMPLE  TABLES 


TOOTH  PROPORTIONS 

Circular 
pitch 

Whole 
depth 

Thickness  of 
tooth 

Equivalent  diametral 
pitch 

GEAR  PROPORTIONS  AND  SPIRALS 


173 


TOOTH  PROPORTIONS 

Diametral 
pitch 

Whole 
depth 

Thickness  of 
tooth 

Equivalent  circular 
pitch 

• 

A.  Square  and  Hexagon 


TRIGONOMETRY  IN  THE  SHOP 

147.  Some  of  the  problems  in  this  section  may  be  solved 
by  arithmetic,  but  more  easily  by  trigonometry.  Others  can 
be  solved  only  by  trigo- 
nometry, for  which  see 

section  249. 

The  method  of  finding 
by  arithmetic  the  largest 
square  or  hexagon  which 
can  be  milled  on  a  cy- 
lindrical piece  of  work, 
or  what  size  to  turn  up 
a  piece  in  order  to  be 
able  to  mill  a  certain- 
sized  square  upon  it  is 
as  follows  : 

148.  For  a  square. 
The    diagonal   of   a 

square  is  equal  to  the 
diameter  of  the  smallest 
cylinder  upon  which  that  square  can  be  made. 

The  diagonal  of  a  1"  square  is  equal  to  the  square  root 
of  2,  or  1.414". 

Using  this  number  1.414"  as  a  constant  C,  to  find  the 
diagonal  of  any  square  in  inches  or  to  find  the  side  of 


B.  Milling  of  a  Square 
FIG.  100 


FIG.  101.    MILLING  MACHINE 
(By  permission  of  Brown  and  Sharpe  Mfg.  Co.) 


PARTS  OF  THE  MILLING  MACHINE 


1.  Spindle  clutch,  for  driving  arbor. 

(For  driving  arbor  or  collet  hav- 
ing clutch  collar.) 

2.  Spindle  adjusting  knob.    (For  ob- 

taining a  tine  adjustment  of 
spindle  by  hand.) 

3.  Spindle-change  feed  lever.    (Moves 

sliding  quill  gears  for  obtaining 
fast  or  slow  series  of  speeds.) 

4.  Spindle-speed  tumbler  gear,  sliding 

knob.  (For  moving  tumbler  gear 
into  position  to  engage  any  one  of 
the  cone  of  gears.) 


Spindle-speed  tumbler  gear,  self- 
locking  lever.  (Throws  tumbler 
gear  in  and  out  of  mesh  with  any 
of  the  cone  of  gears.) 

Feed  lever.  (Gives  feeds  per  revo- 
lution of  spindle,  or  in  inches  per 
minute.) 

Feed  tumbler  gear,  sliding  knob. 
(For  moving  tumbler  gear  into 
position  to  engage  any  one  of  the 
cone  of  gears.) 

Feed  tumbler  gear,  self-locking 
lever.  (Throws  tumbler  gear  in 


174 


and  out  of  mesh  with  any  of  the        29. 
cone  of  gears.) 
9.  Feed  -  changing    lever.      (Upper.) 

(Moves  sliding  quill  gears  to  ob-        30. 
tain  a  fast  or  slow  series  of  feeds.) 

10.  Feed -changing    lever.      (Lower.) 

(Moves  sliding  quill  gears  to  ob-        31. 

tain  an  additional  series  of  fast 

and  slow  feeds.)  32. 

11.  Feed  reverse  lever.  (Starts,  stops, 

and  reverses  all  automatic  table       33. 
feeds.) 

12.  Transverse-feed  trip  lever.    (Con- 

trols automatic  transverse  move-        34. 
ment  of  table.) 

13.  Vertical-feed  trip  lever.  (Controls 

automatic  vertical  movement  of       35. 
table.) 

14.  Longitudinal-feed  trip  lever.  (Con-        36. 

trols  automatic  longitudinal 
movement  of  table;  also  re- 
verses same.)  37. 

15.  Transverse-feed  handwheel.    (For 

use  when   adjusting  or  feeding       38. 
table  by  hand.) 

16.  Vertical-feed  handwheel.  (For  use 

when  adjusting  or  feeding  table        39. 
by  hand.) 

17.  Table  quick-return  handle.     (For        40. 

use  when  adjusting  or  feeding 
table  by  hand ;    also  furnishes        41. 
means  for  quick  return  of  table.) 

18.  Transverse-feed  clutch  knob.  (For 

throwing  handwheel  in  and  out        42. 
of  action.) 

19.  Vertical-feed  clutch  knob.     (For 

throwing  handwheel  in  and  out 

of  action.)  43. 

20.  Longitudinal  -  feed    safety    stops. 

(To  prevent  damage  to  longitu- 
dinal-feeding mechanism  in  case 
the  trip  dog  is  on  the  wrong  side        44. 
of  the  trip  plunger.) 

21.  Longitudinal-feed  trip  dog.    (Ad- 

justable. Throws  feed  out  of  ac- 
tion at  any  desired  point.) 

22.  Transverse-feed  trip  dogs.  (Adjust-       45. 

able.   Throws  feed  out  of  action 

at  any  desired  point.)  46. 

23.  Vertical-feed  safety  stop.   (To  pre- 

vent damage  to  vertical-feeding       47. 
mechanism  when  the  trip  dog  is 
on  the  wrong  side  of  the  trip 
plunger.) 

24.  Vertical -feed  trip  dog.   (  Adjust-       48. 

able.  Throws  feed  out  of  action 
at  any  desired  point.) 

25.  Trip-stop  pin  knob.   (Allows  table        49. 

to  be  fed  only  in  the  desired 
direction.)  50. 

26.  Adjusting  screw.    (For  adjusting 

length  of  feed  chain  for  wear.)  51. 

27.  Arm-clamping  lever.  (Clamps  over- 

hanging arm  at  two  places  by  one 
movement.)  52. 

28.  Friction  -  clutch    lever.      (Short.) 

(Used    for    throwing    clutch    in 
driving  pulley  in  and  out  when        53. 
changing  spindle  speeds.) 

175 


Friction  clutch  lever.  (Long.) 
(Used  for  stopping  or  starting 
machine.) 

Oil  pocket  for  quick  oiling.  (De- 
livers oil  to  all  bearings  in 
spindle  speed  and  feed  cases.) 

Pad  for  motor  bracket.  (Facilitates 
application  for  motor  drive.) 

Spiral  head.  (For  use  in  indexing 
and  the  cutting  of  spirals,  etc.) 

Footstock.  (For  supporting  outer 
end  of  work  when  using  spiral 
head.) 

Rapid  indexing  plate.  (For  rapid 
indexing  of  numbers,  2,  3,  4,  6,  8, 
12,  and  24.) 

Work  driver.  (For  driving  work 
when  using  centers.) 

Footstock  center  adjusting  knob. 
(For  adjusting  center  in  relation 
to  work  arbor.) 

Arbor  yoke.  (Intermediate  support 
for  arbor.) 

Graduations  on  saddle.  (Provides 
means  for  setting  table  to  any 
required  angle.) 

Spiral-head  pan.  (To  hold  spiral 
head  when  not  in  use.) 

Raising  block.  (To  set  spiral  head 
in  any  angle  or  position  on  table.) 

Spiral-head  center.  (For  use  in  dif- 
ferential indexing  for  all  num- 
bers from  1  to  382.) 

Collet.  (To  provide  means  for  hold- 
ing shanks  of  cutters  and  arbor 
with  smaller  tapers  than  spindle 
of  machine.) 

Spiral-head  change  gears.  (Used 
^to  obtain  different  leads  with 
the  spiral  head,  and  differential 
indexing.) 

Table  stops.  (To  govern  travel  of 
table  when  setting  by  hand ;  al- 
so to  form  a  positive  lock  when 
transverse  and  vertical  feeds  are 
not  in  use.) 

Knock-out  rod.  (Removes  arbors 
and  collets  from  spindle.) 

Center  rest.  (To  support  work  car- 
ried by  spiral  heads.) 

Bushing  with  center.  (For  sup- 
porting outer  end  of  arbor.  In- 
terchangeable with  bushing  in 


Arbor-holding  nut.     (For  locking 
arbor  or  collet  and   clutch   to 


Index  plates.  (Used  for  differen- 
tial and  ordinary  indexing.) 

Thread  guard.  (To  protect  thread 
on  end  of  spindle.) 

Chuck.  (For  holding  round  stock 
either  in  machine  or  spiral-head 
spindle.) 

Machine-spindle  chuck  plate.  (To 
hold  chuck  when  used  on  ma- 
chine spindle.) 

Vise.   (For  holding  work.) 


176  SHOP  PROBLEMS  IN  MATHEMATICS 

the  square  when  the  diagonal  is  given  we  have  the  follow- 
ing rule  : 

The  diagonal  of  any  square  is  equal  to  the  side  of  that 
square  multiplied  by  C,  or  the  side  of  any  square  is  equal  to 
its  diagonal  divided  by  C.  ^ 

149.  For  a  hexagon. 

The  distance  across  the  corners  of  a  hexagon  is  equal  to  the 
diameter  of  the  smallest  cylinder  upon  which  that  hexagon 
can  be  made. 

The  distance  across  the  corners  of  any  hexagon  is  equal  to 
twice  its  side. 

By  a  simple  proposition  in  geometry  it  can  be  proved 
that  the  distance  across  the  corners  of  a  hexagon  is  equal 

2 
to  the  distance  across  the  flats  multiplied  by  —i=,  or  1.156. 

Letting  C'  stand  for  this  value  1.156,  we  have  the  rule: 

The  diameter  of  the  smallest  cylinder  upon  which  a  re- 
quired hexagon  can  be  made,  is  equal  to  the  product  of  the 
distance  across  the  flats  and  the  constant  C'. 

In  order  to  find  the  length  of  the  side  of  a  square  or 
the  distance  across  the  flats  of  a  hexagon  which  can  be 
milled  upon  any  given  cylinder,  divide  its  diameter  by  C 
or  C". 

PROBLEMS 

1.  Denoting  the  side  of  the  square  by  s,  the  diagonal  by 
d,  and  the  constant  by  C,  write  a  formula  for  d  in  terms  of 
s  and  C. 

2.  Denoting  the  side  of  a  hexagon  by  h,  the  distance 
across  the  flats  by  I,  and  the  constant  by  C',  write  a  formula 
for  h  in  terms  of  I  and  C'. 

3.  A  shaft  is  .785"  in  diameter.    What  is  the  largest 
square  that  can  be  milled  upon  it  ? 


GEAR  PROPORTIONS  AND  SPIRALS 


177 


4.  How  wide  would  the  flats  (parallel  sides)  of  a  hexag- 
onal head  be  when  milled  on  a  shaft  .895"  in  diameter? 
What  would  be  its  distance  across  the  flats  ? 

5.  A  square  .687"  on  a  side  must  be  milled  on  a  shaft. 
What  is  the  smallest  diameter  to  which  the  shaft  may  be 
turned  ? 

6.  A  f "  square  is  to  be  milled  on  a  |"  shaft     How  many 
thousandths  must  the  shaft  be  reduced  on  each  side?    (See 
Fig.  100.) 

7.  One  cut  on  all  sides 
of  B  (Fig.  100)  has  been 
taken  and  the   work  now 
measures  .733".  How  many 
thousandths   should   the 
table   of   the  milling   ma- 
chine be  raised  in  order  to 
finish  the  square  to  f "  ? 

When  it  is  necessary  to  drill  FlG-  102.   JIG.  PLATE 

holes  with  great  accuracy  in  a 

piece  of  work  which  cannot  be  swung  in  a  lathe,  a  milling  machine 
may  be  used. 

Various  methods  of  measuring  the  distances  for  the  location  of 
these  holes  may  be  employed.  The  longitudinal  and  transverse  feeds 
of  the  milling-machine  table  have  dials  graduated  to  thousandths  of 
an  inch,  and  a  reasonable  degree  of  accuracy  in  measuring  movements 
of  the  table  can  be  obtained  by  use  of  these  dials.  Also  by  knowing 
the  number  of  threads  per  inch  of  the  feed  screws,  dials  may  be  devised 
for  special  work  done  in  large  quantities.  (See  later  examples.) 

The  longitudinal  feed  of  the  table  is  in  the  direction  of  the  length 
of  the  table ;  the  transverse  feed  is  at  right  angles  to  the  length  of 
the  table. 

8.  Holes  A  and  B  (Fig.  102)  are  to  be  drilled  on  the 
milling  machine.    In  order  to  drill  A,  how  much  movement 
of  the  table  will  there  be  in  each  direction  after  drilling  B  ? 


178 


SHOP  PROBLEMS  IN  MATHEMATICS 


3-37 


FIG.  103.    JIG  PLATE 


9.  Find  table  movements  for  drilling  A,  B,  and  C  when 
angle  CBA  is  100°  (Fig.  103). 

10.  Five  holes,  one  at  every   corner  of  a  pentagon   of 
the  size  shown  in  Fig.  104,  are  to  be  drilled  on  a  milling 

machine.  Starting  from  A, 
give  the  lateral  and  vertical 
movements  in  thousandths 
of  an  inch  of  the  milling- 
machine  table  for  all  of 
these  holes. 

11.  Two  plugs  are  located 
3.773"  apart,  measuring  to 
the  outside  of  the  plugs. 
If  plug  A  (Fig.  105)  is  .375" 
in  diameter  and  B  is  .625",  what  is  the  exact  distance  be- 
tween their  centers  ? 

12.  Given  two  plugs  by  which  to  locate 
the    centers    of   two    holes    1.987"   apart. 
What  should    they  measure    over  all,  if 
the  plugs  are  f"  in  diameter? 

13.  It  is  desired  to  locate  points  A,  B, 
C  accurately,  as  shown  in  Fig.  106.    Select 
diameters  for   the  plugs  at  A,  B,  and   C 

such  that  when  they  are  tangent  A  to  C  and  C  to  B,  the 
correct  distances  1.975"  and  1.365" 
will  be  given.   Distance  AB  is  first 
obtained  as  in  Problem  12. 

14.  A  disk  is  10|"  in  diameter.    Flo<  105.  SPACING  HOLES 
Find  the  distance  necessary  to  set 

a  pair  of  dividers  in  order  to  space  off  7  sides ;  8  sides ; 
10  sides  ;  13  sides. 

15.  Six  holes,  one  in  every  corner  of  a  hexagon,  are  to 
be  drilled  on  the  milling  machine.    The  distance  across  the 


FIG.  104. 
PENTAGON 


GEAR  PROPORTIONS  AND  SPIRALS 


179 


flats  of  the  hexagon  is  1.25".    Find  the  lateral  and  vertical 
movements  of  the  milling-machine  table  for  every  hole. 

16.  The  vertices  of  an  equilateral  triangle  are  to  be  lo- 
cated by  3  plugs,  each  measuring  .750"  in  diameter.  The 
sides  of  the  triangle  are  to 
be  4.75".  What  distance  over 
all  must  be  measured  on  the 
plugs  ? 


^'"y\ 


Vvt 


B 


-2.365 
FIG.  106.    JIG  PLATE 


-475'- 
FIG.  107.    ACCURATE  SPACING 


17.  It  is  desired  to  turn  up  a  plug  which  will  just  touch 
the  plugs  in  Problem  16  when  they  are  correctly  located. 
What  should  be  its  diameter  ?    (See  Fig.  107.) 

18.  An  eccentric  is  to  give  a  rocker  arm  A  (Fig.  108)  a 
motion  of  15°.    If  the  throw  of  the  eccentric  is  6",  how 
long  must  the  arm  be  ? 

19.  A  shaft  is  to  have  an  oscillating  motion  of  12°.    If  the 
arm  on  the  shaft  is  5"  long,  what  is  the  throw  of  the  eccentric? 


FIG.  108.    ECCENTRIC  AND  ROCKER  SHAFT 


20.  A  rocker  arm  is  10"  long,  and  a  Second  arm  is  7"  long, 
to  which  the  eccentric  rod  is  attached  (see  Fig.  108).  What 
throw  must  an  eccentric  have  to  move  the  end  of  the  first 
arm  3"?  To  what  angle  would  that  be  equal  ? 


180 


SHOP  PROBLEMS  IX  MATHEMATICS 


21.  Allowing  at  least  1"  of  metal  around  the  eccentric 
shaft  which  is  2"  in  diameter,  how  large  must  the  eccentric 
be  to  meet  the  conditions  in  Problem  20  ? 

22.  A  slide  valve  has  a  motion  of  3".    What  throw  must 
an  eccentric  have  to  produce  this  motion  through  a  bell 
crank  one  arm  of  which  is  7|"  and  the  other  2|-",  if  the 
valve  is  connected  to  the  long  arm  ? 

SPIRALS 

150.  Spirals.    The  spiral  as  used  in  the  machine  shop  is, 
correctly  speaking,  a  helix,  but  in  the  following  problems 
we  shall  follow  the  common  usage. 

151.  Angle  of  spiral  to  axis  of 
work.    The  angle  which  the  spiral 
makes  with  the  axis  of  the  work 

is  easily  found.  Take,  for  example,  JJ 

the  angle  which  a  thread  makes  § 

with  the  axis  of  the  cylinder  upon  Jj 
which  it  is  cut. 


|*-Lead-»| 
FIG.  109.    ANGLE  OF  SPIRAL 

Lay  off  upon  one  side  of  a  right  angle  to  any  convenient 
scale  a  distance  equal  to  the  lead  of  the  spiral ;  upon  the 
other  side  of  the  right  anglej^y  off  a  distance  equal  to  the 
circumference  of  the  cylinder.  Connect  the  ends  of  these 
lines.  The  angle  a  (Fig.  109)  is  the  angle  which  the  spiral 
makes  with  the  axis  of  the  work.  This  is  useful  in  showing 


GEAR  PROPORTIONS  AND  SPIRALS  181 

at  what  angle  the  side  of  the  tool  should  be  ground  for 
clearance  in  cutting  the  thread.  Angle  /3  gives  this  clear- 
ance. It  applies  to  the  angle  which  the  flute  of  a  drill  makes 
with  the  axis  of  the  drill,  and  is  useful  in  setting  the  milling- 
machine  table  for  cutting  the  spiral  of  drills  and  cutters. 
For  the  definition  of  lead,  see  section  114. 

PROBLEMS 

1.  A  £"  twist  drill  has  a  spiral  of  1  turn  in  7.58".   What 
angle  does  the  edge  of  the  flute  make  with  the  axis  of 
the  drill  ? 

2.  A  1-J-"  twist  drill  has  a  spiral  of  9.759"  to  one  turn. 
What  angle  does  the  edge  of  the  flute  make  with  the  axis 
of  the  drill  ? 

3.  Two  spiral   gears  mesh  at  an  angle  of  75°.    Their 
velocity  ratio  is  7  to  8  and  the  angles  of  the  spiral  gears 
are  proportional  to  the  ratios.    What  are  the  angles  ? 

I 


i 

FIG.  110.    SHOWING  MILLING-MACHINE  TABLE 

4.  A  shaft  3"  outside  diameter  has  a  square  thread  •£•" 
deep.    What  angle  does  the  top  edge  of  the  thread  make 
with  the  axis  of  the  shaft  ?    What  angle  does  the  bottom 
of  the  thread  make  ? 

5.  Work  out  a  formula  for  the  angle  of  a  spiral. 

6.  The  shafts  of  two  spiral  gears  which  mesh  together 
make  an  angle  of  90°.    If  both  gears  are  equal,  what  angle 
does  the  teeth  make  with  the  shaft  ? 


182  SHOP  PROBLEMS  IN  MATHEMATICS 

7.  What  spiral  (number  of  turns  to  one  inch)  would  this 
angle  give  on  a  gear  with  a  pitch  diameter  equal  to  3"? 
2"?  1"? 

8.  A  f  "  drill  is  to  be  milled  with  a  spiral  of  6.5"  to  one 
turn.    At  what  angle  must  the  milling  table  be  set  ? 

9.  A  1"  drill  has  a  spiral  of  9"  to  one  turn.    What  angle 
does  its  flute  make  with  the  axis  of  the  drill  ? 

10.  A  |"  drill  is  cut  with  the  table  set  at  20°.    What  is 
the  lead  of  the  drill  ?    What  are  also  the  leads  in  inches 
of  drills  of  the  following  diameters  :  £",  f  ",  f "  ? 

11.  It  is  desired  to  have  the  angle  of  the  teeth  of  a  3" 
spiral  mill  make  about  15°  with  its  axis.    What  should  be 
its  lead  ? 

12.  The  milling  machine  is  geared  to  cut  an  8"  spiral.   If 
the  teeth  are  cut  on  a  1"  cylinder,  what  angle  will  they 
make  with  the  center  line  ?  what  angle  for  a  1£"  cylinder? 
2"?  3"?  4"? 

13.  In  a  worm  and  worm  wheel  the  worm  has  a  lead  of 
i"  and  its  pitch  diameter  is  1J".    At  what  angle  across  the 
face  of  the  worm  wheel  will  the  teeth  be  cut  ? 

Make  a  sketch  of  a  worm  and  worm  wheel.  Notice  the  relations 
between  the  angle  of  the  worm  and  the  angle  of  the  teeth  on  the  face 
of  the  worm  wheel. 

14.  A  worm  has  a  pitch  diameter  of  1"  and  a  lead  of  J". 
What  is  the  angle  between  the  teeth  of  the  worm  wheel 
and  its  axis  ? 

15.  A  worm  has  a  lead  of  0.333"  and  a  pitch  diameter  of 
1-J-".    At  what  angle  must  the  milling  table  be  set  in  order 
to  gash  out  the  teeth  in  the  worm  wheel  ? 

16.  What  is  the  worm-wheel  angle  for  a  worm  of  6  turns 
per  inch,  the  pitch  diameter  being  1J"  ?    What  for  a  worm 
of  5  turns  ?  4  turns  ?   3J  turns  ? 


CHAPTER  XII 
THE  UNIVERSAL  GRINDER  AND  THE  GAS  ENGINE 

THE  GRINDER 

152.  The  angle  of  clearance.  In  backing  off  the  edges  of 
a  cutter  or  reamer  the  edge  makes  a  certain  angle  a  with  the 
tangent  to  the  cutter  at  the  cutting  point  (see  Fig.  111). 


Tanyent   fo  -J      \  ^ 

Cutter 

FIG.  111.    BACKING-OFF  OF  CUTTER 

This  angle  may  be  more  or  less  than  7°,  according  to  the  work 
on  which  the  reamer  is  to  be  used.  The  angle  A  may  be 
ground  by  the  face  of  the  grinding  wheel,  and  for  approxi- 
mate results  we  may  consider  that  the  tangent  line  T  and 
the  face  of  the  wheel  coincide  for  the  short  distance  b. 

153.  Application.  If  it  is  desired  to  back  off  a  cutter  to  a 
certain  number  of  degrees,  say  7°,  with  a  wheel  5"  in  diam- 
eter, a  formula  can  be  derived  which  will  give  the  distance 
D,  or  the  horizontal  distance  of  the  center  of  the  cutter  from 
the  center  of  the  grinding  wheel. 

183 


184  SHOP  PROBLEMS  IN  MATHEMATICS 

In  Fig.  Ill  if  we  may  consider  tangent  T  and  circle  G  to 
coincide  for  the  width  of  the  land  to  be  ground,  then  the 
angle  a  =  a'.  Hence,  by  trigonometry,  D  =  r  sin  a.  Since  r 
and  a  are  given,  sin  a  can  be  found  in  Table  I  and  D  can 
be  calculated. 

EXAMPLE.  Find  the  distance  D  for  a  grinding  wheel  5"in 
diameter. 

D  =  2.5  x  sin  a  =  2.5  x  .122  =  .305". 

PROBLEMS 

1.  Find  the  value  of  D  for  grinding  a  clearance  of  7°  with 
a  wheel  6"  in  diameter;  with  a  wheel  2"  in  diameter;  3^"; 
4";   8". 

2.  Find  the  values  of  D  for  a  5°  angle 
of  clearance,  using  the  same  wheels  as  in 
Problem  1. 

3.  Find  the  values  of  D  for  a  10°  angle 

of  clearance,  using  the  same  wheels  as  in      FlG  112    pISTON 
Problem  1.  RING 

GAS  ENGINE 

154.  The  following  problems  are  of  general  interest ;  they 
occur  in  the  design  of  a  gas  engine  and  are  simple  enough 
to  permit  of  a  place  in  this  book. 

PROBLEMS 

1.  The  ends  of  a  piston  ring  are  intended  to  overlap 
each  other  £"  (Fig.  112)  to  prevent  leakage.    To  what  diam- 
eter should  a  ring  be  turned  so  that  when  sawed  apart 
and  lapped  the  outside  diameter  will  be  3^"  ? 

2.  If  the  clearance  space  of  a  steam  engine  holds  1^  gal. 
of  water,  what  is  the  clearance  in  cubic  inches  ?    What  is 
the  per  cent  of  clearance  if  the  cylinder  is  8"  x  10"? 


UNIVERSAL  GRINDER  AND  GAS  ENGINE      185 

3.  If  the  clearance  space  of  a  gas  engine  holds  11  cu.  in. 
of  water,  the  diameter  of  the  cylinder  is  3\",  and  the  stroke 
3^",  what  is  the  per  cent  of  clearance  ? 

4.  A  water  jacket  of  a  gas  engine  is  f "  wide,  4£"  long, 
and  has  an  inside  diameter  of  4f  ".    How  many  quarts  will 
it  hold?        . 

5.  Find  the  total  area  of  the  valve  seat  in  Fig.  113. 

6.  If  the  total  pressure  exerted  by  a  spring  upon  stem  A 
is  93  lb.,  what  pressure  per  square  inch  is  exerted  on.  the 
valve  seat  ? 

7.  The  maximum  pressure   ex- 
erted  by  the  gas  in  a  gas  engine 
is  about  270  lb.  per  square  inch. 
What  is  the  total  pressure  on  top 

of  the  valve  whose  diameter  is  If"? 

FIG.  113.    VALVE  AND  SEAT 

8.  What  area  of  opening  would 

be  obtained  by  lifting  the  valves  * "  ?   (See  Fig.  113.) 

9.  At  what  velocity  must  gas  enter  a  cylinder  4J"  in 
diameter  by  4i"  long,  and  having  a  valve  as  shown  in 
Fig.  113  in  order  to  fill  it  in  .037  of  one  second  ? 

10.  The  mean  effective  pressure  in  a  gas  engine  is  75  lb., 
the  stroke  4£",  the  cylinder  diameter  4J",  and  the  number  of 
power  strokes  per  minute  400.    Find  the  horse  power  from 

Pi  A  N 

the  formula  HP  =  where  P  is  the  mean  effective 

ooUUU 

pressure,  I  the  length  of  stroke,  A  the  area  of  the  cylinder, 
and  N  the  number  of  power  strokes. 


REVIEW  OF  CALCULATION  WITH 
SHORT  METHODS 

CHAPTER  XIII 

FRACTIONS 

155.  Addition  of  fractions. 
EXAMPLE  1.  Add  ^  and  T\. 

TV  +  fV  =  tt  =  £ 
EXAMPLE  2.  Add  f  and  |. 

t  +  I  =  A  +  if  ^  if  =  1TV 
EXAMPLE  3.  Add  the  mixed  numbers  3f  and  4f . 
3f  +  4*  =  3if  +  4H  -  7|f  *=  8/T. 

ORAL    EXERCISE 

!-  1  +  i-  6.  A  +  A. 

2.  §  +  J.  7.  i  +  A- 

3.  |  +  A-  8.  1\  +  4§. 

4.  A  +  J.  9.  2^  +  31. 

5.  A  +  |.  10.  1J  +  4f  +  2J. 

WRITTEN   EXERCISE 

!•  T5ff  +  |  +  J.  3.  2|  +  3}  +  2/¥. 

2-  ^  +  A  +  I-  4-  5f  +  3|  +  7-^. 

5-  A  +  2i-  +  f 

186 


FRACTIONS  187 

6.  How  wide  a  floor  space  can  be  covered  with  5  boards 
whose  widths  are  respectively  101",  12¥",  9|",  6|",  and  4£"? 

7.  In  laying  pipe  a  plumber  uses  4  pieces  measuring 
respectively  5'  3J",  4'  2|",  10'  3",  and  3'  7f".     Find  the 
total  length  of  the  pipe. 

156.  Subtraction  of  fractions. 
EXAMPLE  1.  From  £  subtract  J. 

5  1    4    2 

I   ~~  ff   —    f   —   3- 

EXAMPLE  2.  Subtract  j\  from  £. 

I  -  T3<r  =  if  -  A  -  iJ- 

EXAMPLE  3.  From  15J  take  9|. 

15  J  -  9|  =  14|  -  9|  =  5f  =  5J. 

ORAL    EXERCISE 

1-  §  -  i-  6'     T50    -  I ' 

2-  I  -  J-  7.   f  r  _  |. 

3.|-|-  8-  M-l- 

9.   3|  -  IJ. 
.  10.   7|  -  3|. 

WRITTEN   EXERCISE 

1.  151 -3J.  5.  26^-18f 

2.  24^  -  19|.  6.  §1  _  lV 

3.  401  -  32§.  7.  1 A  -  /¥. 

4.  31A  -  17J.  8.  5  -  3i|. 

9.  A  tenpenny  nail  is  2|"  long,  a  fourpenny  nail  1§",  and 
a  sixpenny  nail  2".    Find  the  difference  in  lengths  between 

(a)  a  tenpenny  and  a  fourpenny  nail ; 

(b)  a  sixpenny  and  a  fourpenny  nail. 


188  SHOP  PROBLEMS  IN  MATHEMATICS 

10.  A  tenpenny  nail  (see  Problem  9)  is  driven  into  two 
boards  li"  and  1$"  thick,  laid  one  upon  the  other.  How 
far  from  the  lower  surface  is  the  point  of  the  nail  ? 

157.   Multiplication  of  fractions  and  cancellation. 
EXAMPLE  1.  f  of  15  =  ? 

2        5 

|  of  #  =  10. 

EXAMPLE  2.  11  x  35  =  ? 


3 
EXAMPLE  3.  Multiply  jf  by  ^. 

3        7 

15  M  =  SXM  =  ?1 

16  25      JjJ     -JZ0      40 

8        5 

EXAMPLE  4.  Multiply  5f  by  15. 

5f  x  15  =  5  x  15  and  f  of  15 

=  75  and  6 

=  81. 
EXAMPLE  5.  Multiply  3f  by  2|. 

3f  x  2}  =  V  x  -V-  =  W  =  9§i- 
EXAMPLE  6.  Find  the  product  of  |,  ^§,  |,  and 


2       0  4 

EXAMPLE  7.   Divide  24  x  32  x  10  x  45  by  16  x  20  x  9  x  40. 


_ 

x  ?p  x  9  x  £0 


FRACTIONS  189 

EXAMPLE  8.  Find  the  product  of  If  x  3§  x  6J  x  5f. 
3  57 

9      11      n      &_  1155 

6^7?   J  :  7;        5 

ORAL   EXERCISE 
MX  4:  6.    |  of  if. 

2.  f\  x  32.  7.  TV  of  |f 

3.  |  of  £,  8.   |  X  J. 

4.  i|-  of  /T.  9.    |  X  |. 

5.  f\  of  30.  10.    TV  of  f|. 

WRITTEN   EXERCISE 

1.     3?5    X   21.  7.     i  Of  f   Of  §f. 


3.  f  of  §1. 

4.  I  x  H.  9    40x25x16x24 

5.  24JX16.  '27x20x30x12 

6.  16|  x  121.  10.  41  x  71  x  10j|  x  2TV 
11.  How  many  cubic  inches  are  there  in  a  box  111"  x 

13f  "  x  8i",  inside  measure  ? 

158.  Division  of  fractions. 

159.  Reciprocal.    The  reciprocal  of  a  number  is  1  divided 
by  that  number,  e.g.  the  reciprocal  of  5  is  J.   The  reciprocal 
of  |  is  j. 

160.  Rule  for  division.    To  divide  by  a  fraction,  multiply 
by  the  reciprocal  of  the  fraction. 

EXAMPLE  1.    Divide  |  by  |^. 

3 

^10^2^3 
7   '  21      f      W      2 

2 


190  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  2.    |  -*-  3  =  ? 

I  -  3  =  |  -s-  f  =  |  x  i  =  A- 
EXAMPLE  3.    Divide  3£  by  25. 


EXAMPLE  4.  250  -f-  1  =  ? 

*       50       o 
250  -  -  =          x      =  400. 


ORAL    EXERCISE 

1.  What  are  the  reciprocals  of  f  ,  f  ,  3,  £  ? 

2.  §f  H-  5.  5.    1.  -f-  3.  8.    2^  -f-  J. 

3.  81-i.  6>    5  _,  i.  9.    A  ^  r. 

4.  i  -  i.  7.  2^  i  5.  10.  31  4-  8. 

WRITTEN    EXERCISE 

!•   I  -  iV-  6.   ^  -  |. 

2.  150  --5V  7-  4i^6?- 

3-  A  -  if-  s-  II  -  sV 

4.   9^  *  7-  9-   20I  *  18J. 

5-    it  Hh  A-  10.    A  H-  fi- 

ll. The  dimensions  of  a  box,  inside  measure,  are  3'  4", 
2'  6",  and  1'  8".    The  dimensions   of  another  box,   inside 
measure,  are  1'  6",  2',  and  2'  3".    The  volume  of  the  larger 
box  is  how  many  times  as  great  as  that  of  the  smaller  ? 
161.  Decimal  fractions. 
EXAMPLE  1.  Add  3.5,  4.03,  7.275,  and  .2. 
3.5 
4.03 
7.275 

.2 
15.005 


FRACTIONS  191 

EXAMPLE  2.  Subtract  .42  from  5.3. 
5.3 

.42 
4.88 

EXAMPLE  3.  Multiply  3.246  by  2.45. 
3.246 
2.45 
16230 
12984 
6492 
7.95270 

EXAMPLE  4.  Divide  7.285  by  3.5. 

Before  dividing  multiply  dividend  and  divisor  by  such  a  number 
as  will  make  the  divisor  an  integer. 

7.285  -f-  3.5  =  72.85  -s-  35. 


72.85 

70 
285 
280 


35. 


2.08+ 


5 

Check.  3.5  x  2.08  +  .005  =  7.285. 
Check  by  casting  out  nines. 
EXAMPLE  5.  Divide  5.3  by  4.235. 

5.3  -r-  4.235  =  5300  -=-  4235. 


5300. 
4235 


4235. 


1.25+ 


10650 
8470 
21800 
21175 
625 
Check.  4.235  x  1.25  +  .00625  =  5.3. 


192  SHOP  PROBLEMS  IN  MATHEMATICS 

WRITTEN   EXERCISE 

1.  Add  4.25,  7.3,  6.002,  and  5.372. 

2.  From  92.73  subtract  8.673. 

3.  Multiply  4.3126  by  52.7. 

4.  Divide   673.25   by    12.3    correct   to  .three   places   of 
decimals. 

5.  Divide    802.3   by    7.621    correct    to   three   places   of 
decimals. 

6.  The  circumference  of   a   circle  is  3.1416  times  the 
diameter.    If  the  diameter  is  24.5",  find  the  circumference. 

7.  How  many  cubic  inches  in  a  box  whose  inside  dimen- 
sions are  3.4",  8",  1.8"? 

162.  Reduction  of  a  fraction  to  a  decimal. 
EXAMPLE  1.  Reduce  fa  to  a  decimal  correct  to  three  places. 
3.000   28 
28        .107  + 
200 
196 
4 


ORAL    EXERCISE 

Express  as  decimals  : 

1.  f.  3.   i.  5.  &.  7.  |.  9. 

2.  i.  4.   ^.  6.    i.  8.   &.  10. 

WRITTEN    EXERCISE 

Reduce  to  decimals  correct  to  three  places  : 

1.  |.  3.   T\.  5.   2§|.  7.   3ff.  9. 

2.  TV  4.  Jf.  6.  8.34J.  8. 


CHAPTER  XIV 

PERCENTAGE.    AVERAGES.    SQUARE  ROOT 
163.  Important  per  cents. 

J  =  50%  J  : 


i  -  25% 
|  =  75% 


1  =  40% 
J  =  60% 
*  =  80% 

164.  Base.    The  base  is  the  number  of  which  the  per  cent 
is  to  be  found. 

165.  Rate.    The  rate  is  the  number  of  hundredths  to  be 
taken. 

166.  Percentage.    The  percentage  is  the  result  found  by 
taking  a  certain  per  cent  of  the  base. 

167.  Finding  the  percentage.    The  percentage  is  found  by 
multiplying  the  base  by  the  rate. 

EXAMPLE.  4  per  cent  of  $200  ==  $200  x  .04  (since  4  per 
cent  =  .04)  =  $8. 

168.  Finding  the  rate.    The  rate  is  found  by  dividing  the 
percentage  by  the  base. 

EXAMPLE.  If  you  take  an  examination  of  8  questions 
and  answer  5  correctly,  you  have  answered  f,  or  62^  per 
cent,  correctly.  In  this  case  8  is  the  base,  62^  per  cent  is 
the  rate,  and  5  is  the  percentage. 

193 


194  SHOP  PROBLEMS  IN  MATHEMATICS 

169.  Finding  the  base.  The  base  is  found  Inj  dividing  the 
percentage  by  the  rate. 

EXAMPLE.  A  man  receives  10  per  cent  increase  in  salary, 
amounting  to  $5  a  month;  i.e.  $5  is  10  per  cent  of  his 
former  salary.  1  per  cent  is  TV  of  $5,  and  100  per  cent 
is  100  x  A  of  $5  =  $50.  This  result.  could  have  been 
secured  more  simply  as  follows  : 

' 


ORAL    EXERCISE 

1.  Denoting  the  base  by  B,  the  rate  by  R,  and  the  per- 
centage by  P,  state  a  formula  for  P  in  terms  of  B  and  R. 

2.  Solve  the  formula  for  B. 

3.  Solve  the  formula  for  R. 

4.  If  you  are  receiving  $100  a  month  and  get  5  per  cent 
increase,  what  will  your  salary  be  next  month  ? 

5.  If  there  are  200  school  days  in  the  year  and  you 
attend  150  days,  what  is  your  rate  of  attendance  ? 

6.  An  engineer  receiving  $100  per  month  has  a  10  per 
cent  increase  in  salary.    What  is  his  monthly  salary  ?    Six 
months  later  he  is  cut  10  per  cent.   What  is  his  final  salary  ? 
Why  is  this  not  the  same  as  at  first,  —  he  has  had  an  increase 
of  10  per  cent  and  a  reduction  of  10  per  cent  ? 

7.  A  surveyor  buying  a  transit  instrument  listed  at  $200 
gets  40  per  cent  discount  for  cash.    How  much  does  he  pay  ? 

8.  52  is  13  per  cent  of  what  number  ? 

9.  27  is  9  per  cent  of  what  number  ? 

PROBLEMS 

1.  Find  17  per  cent  of  $345. 

2.  Find  13J  per  cent  of  $523. 


AVERAGES  195 

3.  If  your  salary  is  $1200  a  year  and  you  receive  an 
increase  of  $225,  what  is  the  per  cent  of  increase? 

4.  $500  is  14  per  cent  of  what  number  ? 

5.  On  January  1, 1908, 1  deposit  $400  in  a  savings  bank 
paying  4  per  cent.    The  bank  adds  the  interest  to  the  prin- 
cipal July  1  and  January  1  of  each  year.    How  much  can  I 
draw  on  July  1,  1910? 

6.  If  a  man's  savings  are  $150  on  a  salary  of  $1100, 
what  per  cent  does  he  save  ? 

7.  The  premium  on  a  life-insurance  policy  for  $2000  is 
$52  per  year.    What  is  the  rate  paid  for  insurance  ? 

8.  If  you  borrow  $100  on  a  note  of  3  months  date.d 
to-day,  and  have  it  discounted  on  the  same  day  at  6  per 
cent,  what  are  the  proceeds  of  the  note? 

9.  $14.50  is  18  per  cent  of  what  number  ? 

10.  What  sum  of  money  must  be  deposited  in  a  savings 
bank  paying  4  per  cent,  to  yield  a  semiannual  income  of 
$1500? 

AVERAGES 

170.  Average.  The  average  of  two  or  more  quantities  is 
the  result  of  adding  the  quantities  and  dividing  the  sum  by 
the  number  of  quantities ;  e.g.  to  average  3,  8,  7,  6,  and  9, 
divide  their  sum,  33,  by  5,  and  the  average  is  6§. 

The  average  may  often  be  found  without  adding  the 
numbers  ;  e.g.  to  find  the  average  of  68,  70,  75,  and  60,  take 
any  convenient  number,  as  60,  for  a  base. 
68-60-    8 
70  -  60  =  10 
75  -  60  =  15 
33 

33  +  4    =  8}. 
60  -f  8J  =  68|,  the  average. 


196  SHOP  PROBLEMS  IN  MATHEMATICS 

Again,  find  the  average  of  75,  90,  85,  and  65.    Taking  70 
for  a  base, 

75  -  70  =  5 
90-70=  20 
85-70=  15 
65  -  70  =  -_6 
35 

35-4    =    8f 
70  +  8|  =  78J 

ORAL   EXERCISE 

Find  the  averages  of  the  following : 

1.  8,  7,  6,  4,  3,  9.  3.  10,  15,  20. 

2.  7,  9,  10,  4.  4.  40,  45,  35,  50. 

5.  A  student  has   marks   on  the  four  quarter  years   as 
follows  :•  75,  85,  68,  70.     Find  the  average. 

6.  A  man  works  8  hr.  Monday,  10  hr.   Tuesday,  9  hr. 
Wednesday,  7  hr.  Thursday,  11  hr.  Friday,  and  10  hr.  Sat- 
urday.   Find  the  average  length  of  his  day  for  the  week. 

PROBLEMS 

1.  At  different  times  during  the  day  an  indicator  shows 
a  pressure  of  steam  of  175,  180,  140,  100,  65,  50,  40,  and 
45.    Find  the  average  or  mean  pressure. 

2.  What  is   the  average  speed  of  a  train  that  travels 
200  yd.  in  10  sec.,  400  yd.  in  15  sec.,  and  75  yd.  in  5  sec.  ? 

3.  The  maximum  temperature  at  a  certain  mountain  resort 
on  seven  successive  days  was  60°,  65°,  68°,  70°,  52°,  66°,  and 
72°.   Find  the  mean  of  maximum  temperature  for  the  week. 

4.  The  minimum  temperature  at  the  same  place  for  the 
seven  days  was  40°,  42°,  48°,  51°,  47°,  50°,  and  52°.    Find 
the  mean  of  minimum  temperature. 


SQUARE  ROOT  197 

5.  Five  planks  measure  respectively  10' 3",  12' 4",  11' 6", 
9'  8",  and  10'  9".    Find  their  average  length. 

6.  A  board  is  18^"  wide  at  one  end  and  11  £"  at  the  other. 
What  is  the  average  width  ? 

171.  Square  root. 

EXAMPLE  1.  Find  the  square  root  of  2025. 

2025(45  Check 

16  452  =  2025 

85J425 
425 

EXAMPLE  2.  Find  the  square  root  of  138,384. 

138384(372  Check 

9  3722  =  138,384 

67)483  (Casting  out  nines) 

469  0=0 
742)1484 
1484 

EXAMPLE  3.  What  is  the  square  root  of  2.3  correct  to 
two  decimal  places  ? 

2.30(1.51+  Check 

J_  1512  +  199  =  23,000 

25  )  130  (Casting  out  nines) 

125  4  +  1  =  5 

301 )  500 
301 
199 

ORAL    EXERCISE 

Find  the  square  root  of : 

1.  121.          3.  64.  5.  225.  7.  .0016.          9.  TV 

2.  36.  4.  144.          6.  .49.  8.  .25.  10.     . 


198  SHOP  PROBLEMS  IN  MATHEMATICS 

WRITTEN   EXERCISE 
Find  the  square  root  of : 
1.  4225.  2.  2809.  3.  3721. 

4.  53361.  5.  21224449. 

Find  the  roots  correct  to  two  decimal  places : 

6.  3.1416.  7.  3.  8.  24.6.  9.  3840.12. 

10.  |$.   (Eeduce  to  decimal  before  finding  square  root.) 
172.  Factoring  method. 

EXAMPLE.    Find  the  square  root  of  the  product  of 
27  x  9  x  32  x  6. 


V27  x  9  x  32  x-  6  =  V33  x  32  x  25  x  2  x  3 

=  V36  X  26  =  33  x  23  =  216. 


ORAL    EXERCISE 


6.   V10  x  5  x  14  x  7. 


7.   V18  x  10  x  5. 
3.   V6  x  3  x  2.  8.   V36  x  49. 


4.   V125  x  5.  9.   V35  x  7  x  2. 


5.   V7  x  14  x  2.  10.   V21  x  14  x  6. 


CHAPTER   XV 

RATIO  AND  PROPORTION 

173.  Ratio.    The  relation  of  one  quantity  to  another  of 
the  same  kind  with  respect  to  magnitude  is  called  their  ratio. 
It  is  expressed  by  dividing  the  first  by  the  second,  e.g. : 

The  ratio  of  $2  to  $3  is  §,  or  2  :  3. 
The  ratio  of  4"  to  8"  is  f ,  or  £,  or  1  :  2. 
The  ratio  of  18  bd.  ft.  to  72  bd.  ft.  is  }|,  or  ',  written 
also  1  :  4. 

174.  Terms  of  a  ratio.    The  numerator  and  the  denomi- 
nator of  the  ratio  are  respectively  the  first  and  second  terms 
of  the  ratio. 

175.  Antecedent.    The  first  term  is  the  antecedent. 

176.  Consequent.    The  second  term  is  the  consequent. 

ORAL    EXERCISE 

Find  the  value  of  x  in  the  following : 

!    *-5       3    E  =  3       5    *_2.     7    *_3        9    8_! 
*'  3  3'  5  5^5  8~4         '  x~ 

8.f-i.    4. >i.   •.£4   iXj.  io.»-A 

4  62  63  x     2  x 

11.  A  door  measures  8'  x  4f.    What  is  the  ratio  of  the 
length  to  the  width  ? 

12.  There  were  25  fair  days  in  November,  the  rest  being 
stormy.    What  was  the  ratio  of  fair  days  to  stormy  ? 

13.  The  ratio  of  the  height  of  a  certain  door  to  its  width 
is  3  :  2.    It  is  4|f  wide.    How  high  is  it  ? 

109 


200 


SHOP  PROBLEMS  IN  MATHEMATICS 


PROBLEMS 

1.  Measure  (Fig.  114)  AD  and  AB 

AD 

and  find  the  ratio  — —  • 
A  -Z> 

2.  Measure  (Fig.  115)  h,  b}  h',  and    *    |_ 

b'.    Find  the  areas  of  the  rectangles 

,„•  ,    ,  .  FIG.  114.  RECTANGLE 

and  their  ratio. 

3.  Does  a  ratio  exist  between  R  and  h  ?   Give  the  reason. 

4.  In  a  tennis  court  (Fig.  116)  find  the  following  ratios : 

Length  of  the  double  court  _  AB 
'  Width  of  the  double  court  ~  17)' 

Length  of  the  single  court        EF 
'  Width  of  the  single  court 


R 


h 


FIG.  115.    RECTANGLES 

Area  of  the  single  court 

(GI ' 

v  '  Area  of  the  double  court 

Length  of  a  service  line,  MN 

Length  of  a  base  line,  AD 
Perimeter  of  the  single  court 

(  f>\      — * 

'  Perimeter  of  the  double  court 

5.  In  a  baseball  diamond  (Fig.  117)  find  the  following 
ratios : 

Distance  from  first  base  to  second 
^  '     Distance  from  first  base  to  third 


RATIO  AND  PROPORTION 


201 


Distance  of  the  pitcher's  box  from  the  home  plate 
Distance  from  the  home  plate  to  second  base 


A 

/  c^ 

B 

E        r 

1                                    F 

R        F 

I 

/ 

s. 

^£ 

•\J 

! 

. 

H 

C 

rr 

r 

J                                    £ 

FIG.  116.    TENNIS  COURT 

C 

Area  of  the  double  tennis  court 

i\  . . 

Area  of  the  baseball  diamond 
Perimeter  of  the  baseball  diamond 
'  Perimeter  of  the  single  tennis  court 


FIG.  117.     BASEBALL  DIAMOND 


6.  In  the  basket-ball  court  (Fig.  118)  find  the  ratios 
Length  of  the  boundary  line  on  one  end 
Length  of  the  side  line 


202 


SHOP  PROBLEMS  IN  MATHEMATICS 


to 


O 


0 


15 


2.' 


FIG.  118.    BASKET-BALL  COURT 

Length  of  a  side  line 
Length  of  a  diagonal 
,  v  Radius  of  the  circle  in  the  center 

(4 


Radius  of  the  circle  from  which  free  throws  are  made 
Width  of  the  players'  lane 


Distance  of  thrower  from  the  basket 
7.  In  a  football  field  (Fig.  119)  find  the  following  ratios : 


*  w 

idth  of  the  field 

\u 

•   330' 

Length  of  the  fiel 

t          IT) 

•! 

"0 
vO 

4 

0  3>     1 

1 

12 

J 

i 

i 

c 

5 

>l 

"^    1 

-j 

q: 

c. 

O 

>o 
(1 

\ 

i 

0 

D 

FIG.  119.    FOOTBALL  FIELD 


RATIO  AND  PROPORTION  203 

(c)  Make  up  three  other  problems  in  ratios  of  lines  on 
the  football  field. 

177.  Inverse  ratio.  The  ratio  4 : 5  is  the  inverse  of  the 
ratio  5  : 4. 

In  Fig.  120  the  lever  F  is  the  fulcrum,  P  the  power,  W 
the  weight,  FB  the  weight  arm,  and  AF  the  power  arm.  It 

P       weight  arm          P       w 

is  a  law  of  physics  that  —  =  -  — ,  or  —  =  —  -In 

W       power  arm  W      p 

other  words,  the  mechanical  advantage  of  the  lever  is  equal 
to  the  inverse  ratio  of  its  arms. 


FIG.  120.    LEVER 


B 


EXAMPLE.    What  power  will  be  neces- 
sary to  balance  a  weight  of  25  Ib.  on  the  .If. 
above  lever,  if  p  =  5  and  w  =  4  ? 
P  _4 

25  ~  5' 


178.   Separating  in  a  given  ratio. 

EXAMPLE.    Divide  $17  between  A  and  B  in  the  ratio  2  :  3. 

Let  2  x  =  the  number  of  dollars  that  A  receives, 

and  3  x  =  the  number  of  dollars  that  B  receives. 

Then  5  x  =  their  sum, 

and  $17  =  their  sum. 

5  a  =  $17. 
x  =  $3f  ,  or  $3.40. 


3  x  =  $10.20. 

Check.    $6.80  +  $10.20  =  $17. 
2 


=      (dividing  by  3.40). 


204  SHOP  PROBLEMS  IN  MATHEMATICS 

ORAL    EXERCISE 

1.  Divide  20  in  the  ratio  2:3;  3:1. 

2.  Divide  18  in  the  ratio  4  :  2. 

3.  Divide  100  in  the  ratio  4:1;  9:1. 

4.  A  board  18  in.  long  is  to  be  divided  in  the  ratio  5  : 1. 
How  far  from  each  end  is  the  point  of  division  ? 

5.  In  water  there  are  2  parts  of  hydrogen  to  1  of  oxygen. 
How  many  parts  of  oxygen  are  contained  in  120  parts  of 
water  ? 

WRITTEN    EXERCISE 

1.  Divide  200  in  the  ratio  of  7  to  1;  5  to  2 ;  2  to  7. 

2.  In  September  the  ratio  of  clear  to  cloudy  days  was 
5:1.    How  many  days  were  clear  ? 

3.  If  a  line  4f  6"  long  is  divided  in  the  ratio  of  5  to  6, 
what  is  the  length  of  each  part  ? 

4.  Divide  a  legacy  of  $25,000  between  A  and  B,  so  that 
their  shares  shall  be  in  the  ratio  of  2  to  5. 

179.  Reducing  ratio  to  a  percentage.  A  ratio  reduced  to  a 
decimal  becomes  a  percentage  ;  e.g.  the  ratio  2:3,  or  f ,  re- 
duced to  a  decimal  becomes  .66f,  or  66f  per  cent. 

PROBLEMS 

1.  In  a  basket-ball  league  of  ten  schools  each  school  plays 
a  game  with  every  other  school.    In  games  already  played, 
if  your  school  has  won  5  games  and  lost  2,  what  is  its  per- 
centage, that  is,  the  ratio  reduced  to  a  decimal,  of  games 
won  to  games  played? 

2.  If  another  school  has  won  2  games  and  lost  7,  find  its 
percentage. 

3.  A  third  school  finishes  the  series,  having  won  5  games 
and  lost  4,  while  a  fourth  school  has  won  4  out  of  7  games 
played.    Which  has  the  higher  percentage  ? 


RATIO  AND  PROPORTION  205 

4.  In  a  class  of  27  students  22  passed  an  examination. 
Find  the  percentage  of  successful  students. 

5.  In  a  class  of  students  23  passed  and  5  failed.    What 
percentage  of  students  passed  ? 

6.  In  Problems   4  and  5  note  which  class  makes  the 
better  record,  and  from  this  determine  whether  a  ratio  is 
increased  or  diminished   by  adding  1  to  both  numerator 
and  denominator. 

7.  Which  is  greater,  —  or >  if  a  <  b  ? 

Reduce  to  a  common  denominator  and  compare  the  numerators. 

8.  Which  is  greater,  ^  or  7 ~ '  if  a<b?    From  this 

o         o  +  c 

determine  whether  a  ratio  with  the  antecedent  less  than 
the  consequent  is  increased  or  diminished  by  adding  a 
number  to  both  numerator  and  denominator. 

9.  Which  is  greater,  -       -  or ?  if  a  <  b  ? 

b  o  —  c 

Ans.  First  ratio  is  >,  — ,  or  <  the  second,  according  as  (6  —  a  —  c) 
is  positive,  zero,  or  negative.  Note  that  in  the  second  case  the  ratio  is 
always  equal  to  1. 

10.  In  a  civil  service  examination,  from  31  candidates 
in  English  24  men  pass ;  from  31  candidates  in  algebra  29 
pass;  from  26  candidates  in  geometry  24  pass.    Compare 
the  records  of  algebra  and  geometry  and  apply  the  formula 
of  Problem  9. 

Here  a  =  24,  &  =  31,  c  =  5,  and  (&  —  a  —  c)  <  5. 

11.  Check  Problem  10  by  finding  the  percentage  of  al- 
gebra and  geometry.    From  this  notice  which  record  would 
have  been  better,  to  have  had  5  more  men  pass  or  to  have 
had  5  fewer  candidates  ? 

12.  In  cases  like  Problem.  10  would  it  be  possible  to  have 
(b  -  a  -  c)  <  0  ? 


206 


SHOP  PROBLEMS  IN  MATHEMATICS 


13.  In  the  ratio  f^f,  which  operation  gives  the  greater 
ratio,  adding  9  to  the  numerator  or  subtracting  9  from  the 
denominator  ?  (Here  (b  —  a  —  c)  <  0.) 

180.  Proportion,  proportionally,  pro  rata.  These  terms  are 
sometimes  used  meaning  "in  the  same  ratio  as";  e.g. 
(1)  Between  the  center  of  the  earth  aiid  its  surface  the 
weight  of  a  body  is  proportional  to  its  distance  from  the 
center  of  the  earth. 

(2)  A  bonus  of  10  per  cent  of  their  salaries  was  given  to 
three  engineers  as  a  reward  for  completing  their  work  ahead 
of  time.  Their  salaries  were  respectively  $2500,  $1500,  and 


FIG.  121.    GRADE 

$1000  per  year.  Their  total  salaries  being  $5000,  the  first 
received  TV  of  $2500,  or  $250;  the  second,  T\T  of  $1500,  or 
$150;  and  the  third,  TV  of  $1000,  or  $100. 

This  may  be  expressed  in  any  of  the  following  ways  : 

(a)  The  amount  each  received  was  proportional  to  his 
salary. 

(b)  The  bonus  was  divided  among  the  engineers  propor- 
tionally, according  to  salary. 

(c)  The  bonus  of  10  per  cent  was  divided   among  the 
engineers.    The  first  received  10  per  cent  of  $2500,  or  $250, 
and  the  others  pro  rata. 

GRADE 

181.  Grade  of  roadbed.    If  a  trolley  roadbed  rises  h  ft.  in 
a  horizontal  distance  of  100  ft.,  the  ratio  -rr  is  the  grade. 


RATIO  AND  PROPORTION 


207 


If  k  =  2',  the  grade  is  T-p-r*  or  2  per  cent.    When  the  rise 
1UU 

is  uniform  and  it  is  not  convenient  to  measure  100  hori- 
zontal feet,  the  following  method  may  be  used  : 


Fi<r.  122.    GRADE 


Measure  any  convenient  plumb  line,  c  (Fig.  122),  and  the 

corresponding  horizontal  distance,  d.   The  ratio  —  is  equal 

h   .  .        h         c 
to  the  ratio  —  ^.e.  — =  -. 

If  the  grade  is  not  uniform,  the  average  grade  may  be 
found  by  taking  a  series  of  measurements,  as  in  Fig.  123. 


FIG.  123.    GRADE 


This  work  is  usually  done  with  a  level  and  rod,  reading 
by  means  of  a  vernier  to  thousandths  of  a  foot.  The  rod 
is  usually  graduated  to  hundredths  of  a  foot.  In  the  follow- 
ing problems,  results  correct  to  hundredths  of  a  foot  are 
sufficiently  accurate  for  this  kind  of  work. 


208  SHOP  PROBLEMS  IN  MATHEMATICS 

PROBLEMS 

1.  What  is  the  grade  if  h  =  5'?  3.5'?  4.25'?  7.75'?  10'? 

.    h         c  100 c 

2.  If  — —  =  -?  show  that  h  =  —   —  • 

100      d  d 

Using  h  =       .      as  a  formula,  find  h  in  the  following 
problems : 

3.  c  =  5',d  =  50'.  5.  c  =  1.4',  d  =  35'. 

4.  c  =  .8',  d  =  40'.  6.  c  =  .42',  <Z  =  70'. 
Find  the  average  grade  of  a  road,  having  given  : 

7.  c1  =  7',  c"  =  4',  c'"  =  6',  d'  =  85',  d"  =  45',  d'"  =  70'. 

8.  c'  =  5.4',  c"  -  5.6',  c'"  =  11.7',  d'  =  90.3',  d"  =  73.5', 
d'"  =  95.8'. 

9.  e'  =  8.42',  c"  =  9.32',c'"  =  6.26',  rf'  =  73.51',  d"  =  60.25', 
d"f  =  82.17'. 

10.  e'  =  7.68',  e"  =  7.53',  c"' =  3.17',  d'  =  27.51',  d"  =  37.26', 
d'"  =  72.18'. 

PROPORTION 

182.  Proportion.   A  proportion  is  an  expression  of  equality 
between  two  ratios ;   e.g.  the  ratio  8  : 12  is  equal  to  the 
ratio  16  : 24,  since  each  expressed  as  a  fraction  reduces 
to  §.  Therefore,  8  : 12  —  16  :  24,  a  proportion  which  is  read, 
"  8  is  to  12  as  16  is  to  24." 

This  proportion  may  be  written  T8^  =  Jf . 

h        c 

In  section  181,  T—-  =  -  is  a  proportion. 
lUU       d 

183.  Extremes  of  a  proportion.    The  extremes  of  a  pro- 
portion are  the  first  and  last  terms ; 

e.g.  if  §  =  |,  2  and  6  are  the  extremes. 

184.  Means  of  a  proportion.    The  means  of  a  proportion 
are  the  second  and  third  terms ; 

e.g.  if  §  =  |,  3  and  4  are  the  means. 


RATIO  AND  PROPORTION  209 

185.  Rules  for  proportion,    (a)   The  product  of  the  means 
is  equal  to  the  product  of  the  extremes  ; 

e.g.  in  §  =  f ,  observe  that  3x4  =  2x6. 

In  -  =  ->  multiplying  by  Id  we  have  ad  =  be. 

Tf  3       5 

If  —  =  —> 

X          1 


and 


—  7x3,  (Product  of  means  =  product  of  extremes.) 

7x3 


5 

Hence  the  rule : 

(#)  In  any  proportion  the  product  of  the  means  divided  by 
the  given  extreme  is  equal  to  the  required  extreme. 

WRITTEN   EXERCISE 

Find  the  value  of  x  in  Exercises  1-8. 

1.  x  :  25  =  13  :  14.  5.   11  :  12  =  18  :  x. 

2.  x  :  7  =  12  :  17.  6.  125  :  x  ==  206  :  305. 

3.  8  :  x  =  5  :  11.  7.  144  :  105  =  x  :  24. 

4.  7:9  =  ^:14.  8.  1003  :  1800  =  27  :  z. 

PROBLEMS 

1.  The  Chicago,  Milwaukee,  &  St.  Paul  R,ailroad  uses  for 
the  prevention  of  scale  in  locomotive  boilers  an  alkaline 
compound  consisting  of  3750  gal.  of  water,  2500  Ib.  of  caus- 
tic soda,  and  1500  Ib.  of  soda  ash.    How  many  pounds  of 
each  chemical  are  necessary  when  a  mixture  composed  of 
1000  gal.  of  water  is  used  ?    How  many  pounds  of  each 
should  be  used  to  make  5000  gal.  of  the  mixture?    How 
many  pounds  of  each  are  necessary  if  1000  Ib.  of  caustic 
soda  are  used  ?  if  2000  Ib.  of  soda  ash  are  used  ? 

2.  A  formula  for  a  blue-print  solution  is  as  follows  :   2  oz. 
citrate  of  iron  and  ammonia,  1.33  oz.  red  prussiate  of  potash, 


210  SHOP  PROBLEMS  IN  MATHEMATICS 

16.5  oz.  of  water.  Find  the  amount  of  other  chemicals,  if 
3  oz.  of  citrate  of  iron  and  ammonia  are  used ;  if  3.5  oz.  of 
the  potash  are  used  j  if  64  oz.  of  the  solution  are  required. 

3.  How  many  ounces  of  pure  silver  must  be  melted  with 
200  oz.  of  silver  800  fine  to  make  a  bar  900  fine  ? 

4.  How  many  ounces  of  gold  must  be  melted  with  10  oz.  of 
gold  14  carats  fine  (if  pure)  to  make  an  ingot  18  carats  fine  ? 

5.  How  many  pounds  of  copper  should  be  melted  in  94.5 
Ib.  of  an  alloy  consisting  of  3  Ib.  of  silver  to  4  Ib.  of  copper, 
so  that  the  new  alloy  shall  consist  of  7  Ib.  of  copper  to  2  Ib. 
of  silver? 

6.  If  pig  iron  contains  93  per  cent  of  pure  iron,  3  per 
cent  of  carbon,  2  per  cent  of  sulphur,  and  the  rest  consists 
of  silicon,  phosphorus,  etc.,  how  many  pounds  of  pure  iron 
are  there  in  3  tons  of  pig  iron  ? 

7.  An  ingot  of  gold  and  silver  weighs  13.20  oz.    What  is 
the  weight  of  each  of  the  two  metals,  supposing  the  gold 
and  the  silver  in  the  ingot  to  have  the  same  value,  and  gold 
to  be  worth  15J  times  as  much  as  silver  ? 

8.  Alcohol  is  received  in  the  laboratory  0.95  pure.    How 
much  water  must  be  added  to  a  gallon  of  this  alcohol  so 
that  the  mixture  shall  be  0.5  pure  ?    (Check  the  result.) 

9.  How  much  water  must  be  added  to  a  5  per  cent  solution 
of  a  certain  medicine  to  reduce  it  to  a  1  per  cent  solution  ? 

10.  How  many  ounces  of  silver  700  fine  and  how  many 
ounces  900  fine  must  be  melted  together  to  make  78  oz. 
750  fine? 

11.  If  bell  metal  is  made  of  25  parts  of  copper  to  11 
parts  of  tin,  find  the  weight  of  each  metal  in  a  bell  weigh- 
ing 1044  Ib. 

12.  Gunpowder  being  composed  of  $•  sulphur,  75  per  cent 
niter,  and  the  balance  charcoal,  how  many  pounds  of  each 
are  contained  in  400  Ib.  of  powder? 


a-x  •*- 


RATIO  AND  PROPORTION  211 

13.  An  architectural  drawing  of  a  house  is  made  to  the 
scale  of  | "  =  1  ft.   Find  the  number  of  square  inches  occupied 
on  the  drawing  by  a  lot  of  land  120  ft.  long  and  80  ft.  wide. 

14.  If  the  heating  surface  of  a  locomotive  boiler  is  1862 
sq.  ft.,  find  the  ratio  of  heating  surface  to  grate  surface 
when  the  grate  measures  9  ft.  6"  by  8  ft. 

15.  In  drawing  a  picture  of  a  tower  which  is  180  ft.  high 
and  32  ft.  in  diameter,  the  diameter  is  to  be  represented  by 
a  line  4"  in  length.    By  how  many  inches  should  the  height 
be  represented  ? 

16.  Find  the  ratio  of 

the  volume  of  a  sphere  6"     A  I ^~i 1  B 

in  diameter  to  the  volume 

of  a  cylinder  6"  in  diam- 

,  „„,  .   ,  FIG.  124.    GOLDEN  SECTION 

eter  and  6"  high. 

17.  A  wooden  pattern  from  which  an  iron  casting  is  made 
weighs  GA  per  cent  as  much  as  the  iron.   If  the  pattern  weighs 
35  Ib.  12  oz.,  how  much  does  the  casting  weigh  ? 

18.  If  §"  on  a  map  corresponds  to  7  mi.  of  a  country,  what 
distance  on  the  map  represents  20  mi.  ? 

186.  The  golden  section.    If  AR  is  divided  at  C  so  that 

—  — — ,  AB  is  divided  in  extreme  and  mean  ratio  and  AC 

is  the  golden  section  between  AB  and  CB.  Denoting  AB  by  a, 
A  C  by  x,  and  CB  by  a  —  x,  show  that  x  =  §  a  (approximately). 
Hence  A  C  and  CB  are  approximately  in  the  ratio  3 : 2. 

The  relation  of  the  golden  section  is  one  of  the  most 
beautiful  relations  of  size  and  was  so  used  by  the  Greeks. 

It  is  a  principle  of  art  that  the  nearer  the  proportions  of 
an  article  approximate  3  : 2,  the  ratio  of  the  golden  mean, 
the  more  beautiful  the  article  is. 

Find  examples  of  this  ratio  in  familiar  objects,  such  as 
the  ratio  of  the  height  of  the  back  of  a  chair  to  the  length 


212 


SHOP  PROBLEMS  IN  MATHEMATICS 


of  the  legs;  the  ratio  of  the  length  of  an  oblong  picture 
frame  to  its  width. 

"  All  good  design  depends  on  order  and  order  is  depend- 
ent on  geometry." 

187.  Gear  of  bicycles.  The  gear  of  a  bicycle  means  the 
diameter  of  an  old-fashioned  large  wheel,  which  in  one 
revolution  would  cover  as  much  ground  as  a  modern  bicycle 
when  the  pedal  makes  one  revolution. 

If  G  is  the  gear,  C  the  number  of  cogs   on  the  crank 

2S  c* 
sprocket,  c  the   number  on  the   rear   sprocket,   G  =  — 

C 

where  28  is  the  standard  diameter  of  the  ordinary  wheel. 

The  standard  number  of  teeth  in  the  front  and  rear  sprockets  is 
as  follows : 

Front  sprocket,  20,  22,  24,  26,  28,  30. 
Rear  sprocket,  7,  8,  9,  10,  11. 

PROBLEMS 

28  C 
1.  Solve  G  = for  C  and  c. 

c 

Fill  out  the  blanks  in  the  following  table,  and  find  C 
or  c,  correct  to  the  nearest  standard  size. 


C 

c 

G 

2. 

20 

7 

3. 

20 

11 

4. 

30 

7 

5. 

28 

9 

6. 

8 

84 

7. 

7 

100 

8. 

24 

70 

9. 

22 

75 

R 


CHAPTER  XVI 

MENSURATION 

188.  Area  of  a  rectangle.    The  area  of  a  rectangle  is  equal 
to  the.  product  of  the  base  and  altitude. 

PROBLEMS 

1.  If  A  stands  for  the  area  of  a  rectangle,  a  the  altitude, 
and  b  the  base,  write  the  formula  for  A  in  terms  of  a  and  b. 

2.  Draw  a  rectangle,  meas- 
ure a  and  b,  and  find  A. 

3.  If  a  =  5", 6  =  81", find ,4.    a 

4.  In  a  tennis  court  a  =  36' 
and  b  =  78'.    Find  A.     (See 

Fig.  116,  sect.  176.)  5     . 

5.  Allowing  10' on  the  side  FIG.  125.    RECTANGLE 
lines  and  15'  on  the  back  lines,  find  the  area  of  the  plot  of 
ground  necessary  for  such  a  court. 

6.  The  diagonals  of  a  rectangle  are  equal.    How  could 
this  principle  be  used  in  determining  whether  the  court  as 
laid  out  was  a  true  rectangle  ? 

7.  In  a  right  triangle  the  square  on  the  hypotenuse  is 
equal  to  the  sum  of  the  squares  on  the  other  two  sides  (see 
sects.  44-46).    If  the  hypotenuse  is  c,  one  side  a,  and  the 
other  b,  write  a  formula  for  c  in  terms  of  a  and  b. 

8.  Using  this  formula,  find  correct  to  inches  the  length 
of  a  diagonal  of 

(a)  the  double  tennis  court ; 

(b)  the  single  court. 

213 


21-4 


SHOP  PROBLEMS  IN  MATHEMATICS 


9.  Is  a  triangle  with  c  =  5,  &  =  3,  ft  =  4a  right  triangle  ? 
with  c  =  10,  a  =  6,  ft  =  8  ? 

10.  How  may  this  triangle  be  used  in  making  the  corner 
of  the  tennis  court  a  right  angle  ? 

11.  Draw  a  plan  of  a  plot  of  ground  to  contain  four 
tennis  courts,  making  what  allowance  you  think  necessary 
for  space  about  the  courts,  specifying 

(a)  the  total  area  of  ground  required  ; 
(ft)   the  length  of  each  line  ; 

(c)  a  method  of  laying  out  the  lines  in  their  order  ; 

(d)  a  method  of  obtaining  a  right  angle  ; 

(e)  a  test  for  the  accuracy  of  the  work. 

12.  Copy  this  figure  and  in  each  rectangle  write  the  area 

in  terms  of  a  and  b. 

13.  From  the  above 
figure  complete  the  fol- 
lowing formula  : 


a 

FIG.  126.    (a  +  6)2 


Find  the  numerical 
area  of  the  above  figure 
by  substituting  in  the 
formula  : 

14.  If  a  =  f,  ft  =  8". 

15.  If  a  =  6'  9",  b  = 
3'  6". 

16.  If    a  =  0",     1  = 
10". 

19.  If  a  =  2'  3",  ft  =  l'4". 

20.  If  a  =  4|",  b  =  4|". 


17.  If  a  =  12',  ft  =  Of. 

18.  If  a  =  |",  ft  =  4J". 

21.  A  baseball  diamond  is  a  square  90  ft.  on  a  side.  Find 
its  area.  What  part  of  an  acre  is  the  diamond  ?  (160  sq.  rd. 
=  1  acre.) 


MENSURATION 


215 


22.  Draw  a  diagram  of  a  baseball  diamond. 

23.  Find  the  distance  from  home  plate  to  second  base 
correct  to  feet  and  inches. 

24.  Applying  the  test  for  the  tennis  court  to  the  form  of 
the  diamond,  what  distances  must  be  equal  in  order  that 
the  diamond  shall  be  a  rectangle  ? 

25.  What  further  test  is  necessary  to  prove  that  the  figure 
is  a  square  ? 

189.  Perimeter.  The  perimeter  of  a  figure  is  the  sum  of 
its  sides. 

The  perimeter  of  a  rectangle  is  twice  the  sum  of  its  base 
and  altitude. 

PROBLEMS 

1.  In  Fig.  125  write  the  formula  for^>,  the  perimeter,  in 
terms  of  a  and  b. 


FIG.  127.    TRIANGLE 


FIG.  128.    EQUILAT- 
ERAL TRIANGLE 


2.  Find  the  perimeter  of  the  single  tennis  court  (Sect.  176). 

3.  Find  the  perimeter  of  the  double  tennis 
court. 

4.  Find  the  perimeter  of  the  baseball  dia- 
mond. 

6.  Write  the  formula  for  the  perimeter,  p, 
of  a  square  in  terms  of  s,  its  side. 

190.  Area  of  a  triangle.    The  area  of  a  tri- 
angle  is  equal  to  one  half  the  product  of  the       ISOSCELES 
base  and  altitude.  TRIANGLE 


216 


SHOP  PROBLEMS  IN  MATHEMATICS 


PROBLEMS 


1.  Denoting  the  area  by  A,  the  base  by  b,  and  the  alti- 
tude by  a,  write  the  formula  for  A  in  terms  of  I  and  a. 


FIG.  130.    RIGHT  TRIANGLE 


FIG.  131.    SCALENE  TRIANGLE 


2.  Draw  a  triangle,  measure  b  and  a,  and  find  A. 

3.  Fill  out  the  blanks  in  the  following  table : 

b  a  A 


1 

25" 

15" 

2 

6'  4" 

6" 

3 

16'  3" 

8'    4" 

4 

45'  7" 

24'  11" 

4.  Find  the  area  of  a  triangular  plot  of  ground  whose 
base  is  50'  10"  and  altitude  15'  6". 

191.  Area  of  an  equilateral  triangle.  The  area  of  an  equi- 
lateral triangle  is  equal  to  the  square  of  a  side  multiplied 
by  the  square  root  of  3  divided  by  4- 


PROBLEMS 

1.  Denoting  the  area  by  A  and  each  side  by  a,  express 
A  in  terms  of  a. 

2.  Draw  an    equilateral  triangle,   measure   a    side,   and 
find  A. 

3.  Find  A  correct  to  two  decimal  places  : 

(a)  Ifa  =  4".  (c)  If  a  =  10'  3". 

(b)  If  a  =  6'  6".  (d)  If  a  =  20'  7". 

(e)  If  a  =  35'  10". 


MENSURATION 


217 


4.  Write  the  formula  for  the  perimeter,  p,  of  an  equi- 
lateral triangle  in  terms  of  a  side,  a. 

5.  With  the  values  given  in  Problem  3,  find  p. 

6.  Find  the  area  of  a  plot  of  ground  in  the  form  of  an 
equilateral  triangle,  if  one  side  measures  75'  9". 

192.  Area  of  any  triangle  in  terms  of  its  sides.  The  area 
of  any  triangle  is  equal  to  the  square  root  of  the  product 
of  5,  s  —  a,  s  —  b,  s  —  c,  where  a,  b,  and  c  are  the  sides  and 
s  is  half  the  perimeter. 

PROBLEMS 

1.  Denoting  the  area  by  A,  write  the  formula  for  A  in 
terms  of  a,  b,  c,  and  s. 

2.  Draw  a  triangle,  measure  a,  b,  and  c,  and  find  A  cor- 
rect to  two  decimal  places. 

3.  Fill  out  the  blanks  in  the  following  table : 

a  b  c  A 


I 

5 

6 

9 

2 

13 

14 

15 

3 

9 

10 

17 

Median 


FIG.  132.    TRAPEZOID 


4.  Find  the  area  of  a  triangular  field  whose  sides  are 
40  rd.,  58  rd.,  and  84  rd. 

5.  Find  a  formula  for  the  pe- 
rimeter, PJ  in  terms  of  a,  b,  and  c. 

6.  With  the  values  given  to  a, 
b,  and  c  in  Problem  3,  find  j9. 

7.  Find  the  length  of  the  side  of  a  plot  of  ground  in  the 
shape  of  an  equilateral  triangle,  to  have  the  same  area  as 
the  plot  in  Problem  4. 

193.  Area  of  a  trapezoid.    The  area  of  a  trapezoid  is  equal 
to  the  product  of  the  altitude  by  one  half  the  sum  of  the  bases. 


218 


SHOP  PROBLEMS  IN  MATHEMATICS 


PROBLEMS 

1.  Denoting  the  area  by  A,  the  bases  by  b  and  b',  and  the 
altitude  by  a,  write  the  formula  for  A  in  terms  of  b,  b',  and  a. 

\-\  Q  One  half  the  sum  of  the  bases 

is  equal  to  the  median. 

2.  Denoting  the  median 
by  m,  write  the  formula  for 
A  in  terms  of  a  and  m. 

3.  Draw  a  trapezoid,  meas- 
ure the  bases  and  altitude, 
and  find  the  area. 

4.  Check  the  preceding 
problem    by   measuring    the 

median  and  using  it  to  find 
FIG.  133.    IRREGULAR  FIELD        , , 

the  area. 

5.  Fill  out  the  blanks  in  the  following  table : 

b  b'  a  A 


c 


1 

V 

4' 

3' 

2 

6" 

8" 

5" 

S 

30  rd. 

40  rd. 

15  rd. 

*• 

1'3" 

2/8// 

18" 

6.  A  field  in  the  form  of  a  trapezoid  has  for  dimensions 
a  =  25  rd.,  b  =  75  rd.,  b'  =  80  rd.    Find  the  area. 

7.  Find  the  length  of  the  side  of  a  square  field,  to  have 
the  same  area  as  that  in  Problem  6. 

194.  Irregular  figures.  Irregular  figures  are  usually  meas- 
ured by  dividing  them  into  triangles  and  trapezoids. 

The  irregular  field  in  Fig.  133  may  be  measured  by  draw- 
ing a  diagonal  DD\  then  the  perpendiculars  AB,  CE,  FG, 
and  HK,  thus  dividing  the  figure  into  four  right  triangles 
and  two  trapezoids  whose  areas  are  easily  computed. 


MENSURATION  219 

PROBLEMS 

1.  Compute  the  area  of  the  field  in  Fig.  133,  if  the  dimen- 
sions  in   rods   are  HK  =  25,  KD  =  5,  ED  —  15,  BA  —  35, 
CE  =  30,  BE  =  40,  ED'  =  10,  FG  =  38,  EG  =  6.   How  many 
acres  ?    (160  sq.  rd.  =  1  acre.) 

2.  Draw  to  scale  an  irregular  figure  of  five  sides,  divide 
it  into  triangles  and  trapezoids,  take  the  necessary  measure- 
ments, and  compute  the  area. 

3.  Do  the  same  for  a  figure  of  six  sides. 

4.  Do  the  same  for  a  figure  of  seven  sides. 

5.  Do  the  same  for  a  figure  of  ten  sides. 

195.  A  polygon  is  a  plane  figure  bounded  by  straight  lines. 
The  triangle  and  quadrilateral  are  polygons. 


FIG.  134.    REGULAR  POLYGONS 

196.  A  regular  polygon  is  a  polygon  that  has  equal  sides 
and  equal  angles. 

The  apothem  of  a  regular  polygon  is  the  line  drawn  from 
the  center  of  the  polygon  perpendicular  to  one  of  the  sides, 
as  h  in  the  figure. 

The  short  diameter  or  the  distance  across  the  flats  of  a 
regular  hexagon  is  the  perpendicular  distance  between  two 
opposite  sides.  It  is  equal  to  the  diameter  of  the  inscribed 
circle  or  to  twice  the  apothem. 

The  long  diameter  of  a  regular  hexagon  is  the  distance 
between  two  opposite  vertices.  It  is  equal  to  the  diameter 
of  the  circumscribed  circle. 

197.  The  area  of  a  regular  polygon  is  equal  to  one  half  the 
product  of  the  apothem  and  perimeter. 


220 


SHOP  PROBLEMS  IN  MATHEMATICS 


PROBLEMS 

1.  Denoting  the  area  by  A,  the  apothem  by  h,  and  the 
perimeter  by  p,  write  the  formula  for  A  in  terms  of  h  and  p. 

2.  Draw  a  regular  hexagon  by  using  the  radius  of  a  circle 
as  a  chord  six  times,  measure  the  apothem  and  side,  and 
find  the  area. 

3.  How  much  surface  is  there  in  the  hexagonal  top  of  a 
jardiniere  measuring  6"  on  a  side  ? 

In  the  case  of  the  regular  hexagon  the  apothem  is  always  -  V3 
where  a  is  the  length  of  a  side. 

4.  From  the  last  problem  show  that  the  area  of  a  regular 

hexagon  in  terms  of  its  side  a  is  A  =  -  -  V3. 

-i 

THE  CIRCLE 

198.  In  the  circle  ABC,  ABC  is  the  circumference,  OC 
a  radius,  AB  a  chord,  DE  a  diameter. 

PROBLEMS 

The  circumference  of  a  circle  is  equal  to  its  diameter  mul- 
tiplied by  3.1416  (—  approximately}. 

1.  Denoting    the    circum- 
ference by  c,   the  diameter 
by  d,  and  3.1416  by  TT,  write 
the  formula  for  c  in  terms  of 
TT  and  d. 

2.  Since  d  =  2  r}  where  r 
is  the  radius,  write  a  formula 
for  c  in  terms  of  TT  and  r. 

Find  c  in    the   following 


FIG.  135.    CIRCLE 
3.  d  =  W".     4.  d  =  7'. 


problems  (use  TT  —  272) : 

6.  r  =  3£".    7.  r=11.7'. 


MENSURATION 


221 


8.  Find  the  circumference  of  a  circular  flower  bed  whose 
diameter  is  20'  6". 

199.  The  area  of  a  circle  is  equal  to  ^  multiplied  by  the 
square  of  the  radius. 

PROBLEMS 

1.  Denoting  the  area  by  A,  272-  by  TT,  and  the  radius  by  r, 
write  (1)  the  formula  for  A  in  terms  of  TT  and  r;  (2)  the 
formula  in  terms  of  TT  and  d,  the  diameter. 

2.  Draw  a  circle,  measure  r,  and  find  A. 
Fill  out  the  blanks  in  the  following  tables : 


3. 


4. 


1 

10" 

2 

2'    6" 

i 

25'    4" 

4 

30'    7" 

6 

50'  11" 

1 

7" 

2 

14" 

3 

60'  6" 

4 

2'  4" 

5 

r  7f" 

5.  A  circular  flower  bed  has  a  diameter  of  25'  8".  Find 
the  area. 

200.  Length  of  an  arc.  An  arc  has  the  same  ratio  to  the 
circumference  that  the  angle  at  the  center  has  to  360°. 

Q/J 

Hence  the  arc  AB  =  ^  -  of  the  circumference 

6bO 

36 


Trd 


PROBLEMS 


1.  Denoting  the  length  of  the  arc  by  I  and  the  number  of 
degrees  in  the  angle  by  N,  write  the  formula  for  I  in  terms 
of  N  and  d. 


222  SHOP  PROBLEMS  IN  MATHEMATICS 

2.  The  minute  hand  of  a  clock  is  4'  long.    How  many 
inches  does  its  extremity  move  in  20  minutes  ? 

3.  Find  the  length  of  that  part  of  a  circular  railway 
curve  with  a  radius  of  one  half  a  mile  that  subtends  an 
angle  of  36°. 

4.  Solve  the  formula  in  Problem  1 
for  N. 

5.  Solve  the  formula  in  Problem  1 
for  d. 

6.  A  pendulum  swings  through  an 
angle  of  36°.    The  end  describes   an 

arc  of  12  in.    Find  the  length  of  the        FlG-  136-   ANGLE 

-,   T  OF  36° 

pendulum. 

201.  Area  of  a  sector.    A  sector  is  that  part  of  a  circle 
which  is  bounded  by  two  radii  and  an  arc. 

In  Fig.  137  A  OB  is  a  sector. 

The  area  of  a  sector  is  equal  to  one  half  the  product  of  its 
radius  by  its  arc.  It  has  the  same  ratio  to  the  area  of  the 
circle  that  the  central  angle  has  to  360°. 

PROBLEMS 

1.  Denoting  the  area  of  the  sector  by  A  and  the  number 
of  degrees  in  the  central  angle  by  N}  write  the  formula  for 
A  in  terms  of  N  and  r. 

2.  A  sector  of  a  circle  has  a  central  angle 
of  27°,  and  the  radius  of  the  circle  is  30'. 
Find  the  area  of  the  sector. 

3.  Find  the  area  of  the  sector  of  the  end        A1f"B 
of  the  circular  head  of  a  tank,  if  the  length  of       FIG.  137. 

the  arc  is  28"  and  the  diameter  of  the  head  of     SECTOR  AND 
.,  ,    .     „.  SEGMENT 

the  tank  is  or. 

202.  A  segment  of    a  circle  is  the   part  of   the  circle 
bounded  by  an  arc  and  its  chord. 


MENSURATION  223 

203.  The  area  of  a  segment  ACE  (Fig.  137)  is  equal  to  the 
difference  between  the  area  of  the  corresponding  sector 
OACB  and  the  area  of  the  triangle  OAB.  These  areas  may 
be  found  by  sections  201  and  190-192. 

In  the  following  problems  it  will  be  useful  to  recall  the  geometrical 
principle  that  in  a  right  triangle  having  one  acute  angle  equal  to  30°, 
the  side  opposite  the  30°  angle  is  one  half  of  the  hypotenuse. 

B 

PROBLEMS 

1.  If   angle  A  =  30°  and  EC  =  1', 
then  AB  =  2'.   AC  can 

be   found   by    sections 

44-46.  /     30' 

2.  If    the   right  tri- 

,  FIG.  138.    30°  RIGHT  TRIANGLE 

angle   has   one  angle 

equal  to  45°,  the  triangle  is  isosceles;  p 

e.g.  if  EC  =  8,  and  Z  A  =  45°,  A  C  =  8. 

AB  may  then  be  found. 

Find  the  area  of  the  segment  in 
each  of  the  following  circles  : 

3.  r  =  10",  Z  0  =  60°.  /^$ 

/\ 


a 


4.  r  =  8",    Z  O  =  90°.  8 

5     r  _  2'  6"   /-  0  =  120°  ^1 

'  ' 


204.  An  annulus  or  ring  is  the  area,  comprised  between 
two  circles  that  have  the  same  center. 

Its  area  is  the  difference  between  the  areas  of  the  two 
circles  and  "may  be  denoted  by 

A  =  TT  (R2  -  r2) 

=  7r(R  +  r)  (R  -  r), 

where  R  and  r  are  the  respective  radii  of  the  two  circles. 


224  SHOP  PROBLEMS  IN  MATHEMATICS 

PROBLEMS 

1.  The  inner  and  outer  diameters  of  an  annulus  are  8" 
and  10"  respectively.    Find  the  area  of  the  annulus. 

2.  What  is  the  area  of  the  annulus  be- 
tween two  circles  9"  and  7"  in  diameter  ? 

3.  A  circular  grass  plot  20'  in  diameter 
is  surrounded  by  a  walk  4'  wide.    Find  the 

area  of  the  walk. 

FIG.  140.  PRISM 

4.  A  circular  pond  75'  in  diameter  is  sur- 
rounded by  a  path  10'  wide.    Find  the  area  of  the  path. 

THE  PRISM 

205.  The  volume  of  a  prism  is  equal  to  the  product  of  its 
base  and  altitude. 

In  the  case  of  the  rectangular  block,  the  volume  is  the 
product  of  the  three  edges  that  represent  the  length,  breadth, 
and  height. 

PROBLEMS 

1.  Denoting  the  volume  by  V  and  the  three  dimensions 
by  a,  b,   and  c,  write  the  formula  for  V  in 

terms  of  a,  b,  and  c. 

2.  The  total  area  of  a  rectangular  block  is 

the  sum  of  the  areas  of  the  six  faces.  FlG-  141. 

Denoting  the  total  area  by  T,  find  T  in  RECTANGULAR 
,  „  ,  BLOCK 

terms  ot  a,  b,  and  c. 

3.  Show  that  the  length  of  a  diagonal  of  a  rectangular 
block  is  Va2  +  b2  +  c2. 

4.  The  volume  of  a  cube  is  equal  to  the  cube  of  an  edge. 

Derive  the  formula  for  the  volume  of  a  cube  from  that 
of  Problem  1  by  making  a  =  b  =  c. 


MENSURATION  225 

5.  Derive  the  formula  for  the  total  area  of  a  cube. 

6.  Derive  the  formula  for  the  diagonal  of  a  cube  from 
Problem  3  by  making  a  =  b  =  c. 

1.  Find  the  volume  of  a  rectangular  block  4"  x  2"  x  8". 

8.  Find  the  total  area  of  the  surfaces  of  this  block. 

9.  What  is  the  length  of  its  diagonal  ? 

10.  Find  the  volume  of  a  cube  6'  on  an  edge. 

11.  Find  the  total  surface  area  of  this  cube. 

12.  Find  the  length  of  its  diagonal.  FlG-  142- 

CUBE 

13.  How  many  gallons  of  water  are  contained 

in  a  tank  in  the  form  of  a  rectangular  parallelepiped  whose 
dimensions  are  15'  x  22'  x  8'  6"  ?    (231  cu.  in.  =  1  gal.) 

14.  How  many  cubic  yards  of  earth  must  be  removed 
in  digging  a  canal  1  mi.  1600  ft.  long,  100  ft.  wide,  and 
18  ft.  deep  ? 

15.  Find  the  volume  of  a  brick  measuring  8"x  4"  x  2". 

16.  A  room  measures  15'  x  20'  x  10'.    How  many  cubic 
feet  of  air  does  it  contain?    How  many  square  yards  of 
surface  on  the  walls  and  ceiling? 

17.  In  the  room  of  Problem  16  find  the  length  of  a  piece 
of  string  that  will  just  stretch  from  one  corner  of  the  ceil- 
ing to  the  opposite  corner  of  the  floor. 

18.  How  long  an  umbrella  will  go  into  a  trunk  measuring 
32"  x  17"  x  21",  inside  measure,  (a)  if  the  umbrella  is  laid 
on  the  bottom  ?  (&)  if  it  is  placed  diagonally  between  oppo- 
site corners  of  the  top  and  bottom  of  the  trunk  ? 

19.  A  railroad  embankment  one  quarter  of  a  mile  long 
has  a  section  in  the  form  of  an  isosceles  trapezoid  (i.e. 
a  trapezoid  having  the  nonparallel  sides  equal)  300'  at  the 
base,  200'  at  the  top,  and  25'  high.    How  many  cubic  yards 
of  earth  does  the  embankment  contain  ? 


226 


SHOP  PROBLEMS  IN  MATHEMATICS 


20.  How  deep  must  a  cistern  be  to  contain  800  gal.,  if  its 
length  and  width  are  respectively  8'  4"  and  4'  6"  ? 

21.  A  cubical  tank  14"  deep,  open  at  the  top,  is  to  be  lined 
with  copper.    How  many  square  inches  of  copper  must  be 
used,  not  allowing  for  overlapping  ? 


FIG.   143.    STEEL  PLATE 

22.  How  many  gallons  of  water  will  the  tank  contain  ? 

23.  A  cubical  box  has  an  edge  measuring  4'  4",  outside 
measure.    Find  the  cost  of  painting  the  outside,  including 
the  top,  at  10/  per  square  yard. 

24.  Find  the  weight  of  a  box  and  lid 
made  of  wood  -J-"  thick,  measuring  on  the 
outside  3f  x  2'  x  2',  if  a  cubic  foot  of  the 
wood  weighs  35  Ib. 

25.  A  steel  plate  2"  thick   is  of  the  D 
form  and  dimensions  of  Fig.  143.    If  a 
cubic  inch  of  steel  weighs  .28  Ib.,  find 
the  weight  of  the  plate. 


FIG.  144.    PYRAMID 


THE  PYRAMID 

206.  A  pyramid  is  a  solid  whose  base  is  a  polygon  and 
whose  sides  are  triangles  having  a  common  vertex. 

The  altitude  of  a  pyramid  is  the  distance  from  the  vertex 
to  the  base. 

The  volume  of  a  pyramid  is  one  third  of  the  product  of  the 
area  of  the  base  by  the  altitude. 


MENSURATION 


227 


207.  A  regular  pyramid  is  a  pyramid  whose  base  is  a  regu- 
lar polygon  and  the  foot  of  whose  altitude  coincides  with 
the  center  of  the  base. 

The  slant  height  of  a  regular 
pyramid  is  the  altitude  of  a  lat- 
eral face  or  the  distance  from  the 
vertex  to  a  side  of  the  base. 

208.  The  lateral  area  of  a  regular 
pyramid  is  one  half  the  product  of 
the  perimeter  of  the  base  and  the 
slant  height. 

209.  The  frustum  of  a  pyramid  is  that  part  of  the  pyramid 
included  between  the  base  and  a  plane  parallel  to  the  base. 

210.  Denoting  the  volume  of  the  frustum  of  a  pyramid  by 
V,  the  altitude  of  the  frustum  by  h,  the  upper  base  by  b',  and 
the  loiver  base  by  b, 


FIG.  145.    REGULAR 
PYRAMIDS 


211.  The  lateral  area  of  the  frustum  of  a  regular  pyramid 
is  equal  to  one  half  the  sum  of  the  perimeters  of  the  upper 
and  lower  bases  multiplied  by  the  slant  height  of  the  frustum. 


PROBLEMS 

1.  Denoting  the  volume  of  the  pyramid  by  V,  the  base 
by  b,  and  the  altitude  by  //,  write  the  formula  for   V  in 
terms  of  b  and  h. 

2.  Denoting  the  lateral  area  of  the  pyramid 
by  Z,  the  perimeter  of  the  base  by  p,  and  the 
slant  height  by  I,  write  the  formula  for  L  in 
terms  of  p  and  I. 

Find  the  lateral  surface  and  volumes  of  regu-       pYRAMID 
lar  pyramids  with  the  following  dimensions  : 

3.  Base  a  square  containing  64  sq.  ft.,  altitude  8  ft. 


FIG.  146. 


228  SHOP  PROBLEMS  IN  MATHEMATICS 

4.  Base  a  square  13"  on  a  side,  altitude  10". 

5.  Base  an  equilateral  triangle  6"  on  a  side,  altitude  8". 

6.  Each  face,  including  the  base,  an  equilateral  triangle 
10"  on  an  edge. 

7.  The  base  a  square  15"  on  a  side,  lateral  edge  16". 

8.  Base  regular  hexagon  6"  on  a 

side,  altitude  8".  /Tl 

9.  Find  the  volume  of  a  pyramid       /     a. 
whose  base  is  a  rhombus  6"  on  a  side,    /(>o'       \ 

and  whose  height  is  6",  if  one  angle   _ 

FIG.  147.    60°  EHOMBUS 
oi  the  rhombus  measures  60  . 

In  a  right  triangle  where  one  acute  angle  is  60°,  the  side  opposite 

Vs 

this  angle  is  equal  to  the  hypotenuse  multiplied  by Thus  in  the 

/7>  2 

figure  a  =  6  x  —  =  3V3. 

10.  The  Great  Pyramid  in  Egypt  is  480f  ft.  in  height, 
and  its  base  is  a  square  measuring  764  ft.  on  a  side.    Find 
its  volume  in  cubic  yards. 

Find  the  volume  of  the  frustums  of  pyramids  having  the 
following  dimensions : 

11.  Lower  base  a  square  4'  on  a  side,  upper  base  2'  on  a 
side,  altitude  3'. 

12.  Lower  base   an  equilateral   triangle  8"  on   a   side, 
upper  base  5"  on  a  side,  altitude  4". 

13.  Find  the  lateral  area  of  the  frustum  of  a  square 
pyramid  the  edge  of  whose  lower  base  measures  10",  the 
edge  of  the  upper  base  7",  and  the  slant  height  6". 

THE  CYLINDER 

212.  A  cylinder  of  revolution  or  a  right  circular  cylinder  is 
the  solid  generated  by  the  revolution  of  a  rectangle  about 
one  of  its  sides  as  an  axis. 


MENSURATION 


229 


FIG.  148. 
CYLINDER 


The  rectangle  ABCD  revolving  about  AB  as   an   axis 
generates  the  cylinder,  as  in  the  figure. 

The  bases  of  the  cylinder  are  evidently  circles. 

213.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the 
product  of  the  base  and  altitude. 

214.  The  lateral  surface  of  a  cylinder  of  revolu- 
tion is  equal  to  the  product  of  the  circumference 
of  the  base  by  the  altitude. 

215.  The  total  surface  of  a  cylinder  of  revolu- 
tion is  equal  to  the  area  of  the  bases  added  to 
the  lateral  area. 

216.  A  frustum  of  a  cylinder  is  the  part  in- 
cluded between  a  base  and  a  plane  oblique  to  the  base  and 
cutting  the  lateral  surface. 

217.  The  volume  of  a  frustum  of  a  cylinder  of  revolution  is 
equal  to  the  product  of  the  base  and  the  altitude  to  the  base 
from  the  center  of  the  oblique  section. 

A  R  -4-  C*  T) 

In  the  figure  the  altitude  is  equal  to  -    — 

A 

218.  The  volume  of  a  hollow  cylinder  is  the 

difference  between  the  cylinders  of  its  out- 
side and  inside  diameters. 

PROBLEMS 


1.  If  V  is  the  volume  of  the  cylinder,  r 
the  radius  of  the  base,  and  a  the  altitude, 

write  the  formula  for  Fin  terms  of  r  and  a.  FIG.  149.  FRUSTUM 

*•'..,,.  OF  A  CYLINDER 

2.  If  L  is  the  lateral  surface  of  the  cylin- 
der, write  the  formula  for  L  in  terms  of  r  and  a. 

3.  Derive  a  formula  for  T,  the  total  surface  in  terms  of 
a  and  r. 

4.  If  V  is  the  frustum  of  a  cylinder,  write  a  formula  for 
V  in  terms  of  r,  h',  and  h"  (see  Fig.  149). 


230  SHOP  PROBLEMS  IN  MATHEMATICS 

Find  the  volumes,  lateral  areas,  and  total  areas  of  cylin- 
ders of  revolution  having  dimensions  as  follows : 

5.  Diameter  of  base  2',  height  18". 

6.  Diameter  of  base  18'  6",  height  50". 

7.  Find  the  volume  of  the  frustum  of  a  cylinder  the 
radius  of  whose  base  is  25"  and  whose  altitude  from  the 
center  of  the  oblique  section  to  the  base  is  30". 

8.  What  is  the  weight  of  a  hollow  cylinder  of  cast  iron 
20'  long  and  1"  thick  whose  outside  diameter  is  3'  ?    (1  cu. 
in.  of  cast  iron  weighs  .26  Ib.) 

9.  How    many   gallons    (231    cu.   in.)    of  water  will    a 
cylindrical  tank  hold  that  is  15'  in  diameter  and  25'  high  ? 

10.  How  much  will  it  cost  to  paint  the  tank  in  Problem 
9  at  15/  per  square  yard  ? 

11.  How  many  square  inches  are  there  in  the  curved 
surface  of  a  wire  -J-"  in  circumference  and  100'  long  ? 

12.  Find  the  number  of  square  yards  in  the  surface  of 
a  smokestack  40'  high  and  3'  in  exterior  diameter. 

13.  How   many  cubic  feet  of  water  are   contained  in 
100  yd.  of  pipe  whose  internal  diameter  is  2"  ? 

14.  If  a  cubic  foot  of  marble  weighs  173  Ib.,  find  the 
weight  of  a  cylindrical  marble  column  20'  high  and  25"  in 
diameter. 

15.  Water  is  poured  into  a  cylindrical  reservoir  25'  in 
diameter  at  the  rate  of  300  gal.  a  minute.    Find  the  rate 
(number  of  inches  per  minute)  at  which  the  water  rises 
in  the  reservoir. 

16.  Find  the  cost  of  digging  a  well  100'  deep  and  5'  6" 
in  diameter  at  an  average  cost  of  $3  per  cubic  yard. 

17.  Find  the  weight  of  a  copper  tube  £•  in.  in  outside 
diameter,  .05  in.  thick,  and  6  ft.  long,  if  1  cu.  in.  of  copper 
weighs  .318  Ib. 


MENSURATION 


231 


18.  The  larger  diameter  of  a  hollow  cast-iron  roller  is 
1'  10",  the  thickness  of  the  metal  is  If",  and  the  length  6'. 
Find  the  weight  of  the  roller. 

THE  CONE 

219.  A  cone  of  revolution  (usually  called  a  cone)  is  the 
solid  generated  by  the  revolution  of  a  right  triangle  about 
one  of  its  legs  as  an  axis. 

In  the  figure  the  re  volution  of  the  triangle 
A  CB  about  A  C  as  an  axis  generates  the  cone. 

The  base  is  a  circle,  the  altitude  is  h, 
and  the  slant  height  is  I. 

220.  The  volume  of  a  cone  is  one  third  the 
product  of  the  base  and  the  altitude. 

221.  The  lateral  surface  of  a  cone  is  equal 
to  one  half  the  product  of  the  circumference 
of  the  base  by  the  slant  height. 

222.  A  frustum  of  a  cone  is  the  part  of 
a  cone  included  between  the  base  and  a 
plane  parallel  to  the  base. 

223.  The  volume  of  a  frustum  of  a  cone  of 
revolution  is  equal  to 


FIG.  150.  FRUSTUM 
OF  A  CONE 


FIG.  151.    CONE 


where  h  is  the  altitude  and  b,  b'  are  the 


bases  of  the  frustum. 

224.   The  lateral  surface  of  a  frustum  of  a 
cone  of  revolution  is  equal  to  one  half  the  product  of  the  slant 
height  by  the  sum  of  the  circumferences  of  the  bases. 

PROBLEMS 

1.  If  V  is  the  volume  of  a  cone  of  revolution,  r  the  radius 
of  the  base,  and  h  the  altitude,  write  the  formula  for  V  in 
terms  of  r  and  h. 


232 


SHOP  PROBLEMS  IN  MATHEMATICS 


2.  Denoting  the  lateral  surface  of  the  cone  by  L  and  the 
slant  height  by  I,  write  L  in  terms  of  r  and  L 

3.  Write  the  formula  for  L,  the  lateral  area  of  a  frustum 
of  a  cone  of  revolution,  in  terms  of  r,  the  radius  of  the  upper 
base ;  •/•',  the  radius  of  the  lower  base ;  and  I,  the  slant  height. 

Find  the  volumes  and  lateral  areas  of  cones  of  revolution 
having  the  following  dimensions  : 


Diameter  of  base 


Altitude 


Slant  height 


4. 

6" 

10" 

5. 

5' 

4' 

6. 

14'  6" 

12'  4" 

7. 

28" 

25" 

8. 

8JW 

6f" 

9.  Find  the  volume  of  the  frustum  of  a  cone  of  revolu- 
tion having  10"  for  the  diameter  of  the  lower  base,  7"  for 
the  diameter  of  the  upper  base,  and  a  height  of  4". 

10.  Find  the  lateral  area  of  the  frustum  of  a  cone  of 
revolution  the  diameters  of  whose  upper  and  lower  bases 
are  4'  and  8'  respectively,  and  whose  slant  height  is  3'  6". 

11.  A  conical  spire  has  a  slant  height  of  60'  and  the 
perimeter  of  the  base  is  50'.    Find  the  lateral  surface. 

12.  Find  the  weight  of  a  solid  circular  cone  of  cast  iron 
whose  height  is  8"  and  the  diameter  of  whose  base  is  5". 

13.  What  length  of  canvas  f  yd.  wide  will  be  required 
to  make  a  conical  tent  measuring  12'  high  and  16'  in  diam- 
eter at  the  base  ? 

14.  Find  the  volume  of  a  cone  with  an  elliptical  base, 
major  axis  5",  minor  axis  3",  and  height  8". 

The  area  of  an  ellipse  is  equal  to  Trafc,  where  a  and  b  are  the 
semiaxes. 


MENSURATION 


233 


15.  A  tin  funnel  is  to  be  constructed  with  the  dimensions 
as  follows  :  10"  in  diameter  at  the  top,  1"  in  diameter  at  the 
bottom,  slant  height  14",  spout  1"  in  diameter  at  the  larger 
end  and  -J-"  at  the  smaller  end,  slant  height  4".    Allowing 
y  on  the  length  and  width  of  each  piece  for  locks,  find 
the  amount  of  tin  required  for  the  funnel. 

16.  A  cone-shaped  hood  for  a  chimney 
is  30"  in  diameter  and  13"  high.  Allowing 
1"  on  the  side  for  locking,  what  surface 
of  metal  is  required  for  the  hood  ? 

17.  A  copper  teapot  is  9|"  in  diameter  FlG  152    SPHERE 
at  the  bottom,  8"  at  the  top,  and  11"  deep. 

Allowing  42  sq.  in.  for  locks  and  waste,  how  much  metal  is 
required  for  its  construction  without  the  cover  ? 

THE  SPHERE 

225.  A  sphere  is  a  solid  generated  by  the  revolution  of  a 
circle  about  a  diameter. 

226.  The  volume  of  a  sphere  is  equal  to 

V  =  -  Trr3,  where  r  is  ike  radius. 


or  V  = )  where  d  is  the  diameter. 

6 

227.   The  surface  of  a  sphere  is  equal  to 


FIG.  153.  SEGMENT 
OF  A  SPHERE 


or  S  = 

228.  A  segment  of  a  sphere  is  the  part  included  between 
two  parallel  planes. 

229 .  The  volume  of  a  spherical  segment  is  -  (Trr2  +  Trr'2)  +  — —  > 
where  h,  r,  and  r1  are  the  dimensions  represented  in  the  figure. 

The  volume  of  a  spherical  segment  is  thus  equal  to  that  of  two 

cylinders  (-Trr2  and  -Trr''2)  and  a  sphere  (— )  •    The  cylinders  have 
\2  2        /  \  6  / 


234  SHOP  PROBLEMS  IN  MATHEMATICS 

a  common  altitude,  -,  and  the  radii  of  their  bases  are  r  and  r'  re- 
spectively.   The  sphere  has  a  diameter  of  h. 

230.  The  volume  of  a  spherical  segment  of  one  base  is  equal 
to  Trk2  I  r  —  - 

231.  A  zone  is  the  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  parallel  planes. 

232.  Its  area  is  2irrh  where  r  is  the  radius  of  the  sphere 
and  h  the  altitude,  as  in  the  figure  above. 

233.  Ring.  If  the  circle  whose  center  is  O  (Fig.  154)  be  re- 
volved about  AB  as  an  axis,  the  figure  formed  is  a  ring.   ED 
is  the  internal  radius,  CD  the  external  A 
radius,  and  OD  the  mean  radius.    The 

ring  may  be  considered  as  a  cylinder 
bent  in  the  form  of  a  circular  arc  till  c 
the  ends  meet.  The  mean  length  of  this 
cylinder  is  evidently  the  circumference 

of  the  circle  whose  radius  is  OD. 

oo/i     mi,  *  '  -i    i      FIG.  154.   CIRCLE  GEN- 

234.  The  area  of  a  ring  is  equal  to 

ERATING    A    RlNG 

the  product  of  the  circumference  of  a 
cross  section  and  the  mean  length.    Denoting  this  area  by 
A}  the  radius  OE  of  the  cross  section  by  r,  and  the  mean 
radius  OD  by  R, 


235.  The  volume  of  a  ring  is  equal  to  the  product  of  the 
area  of  a  cross  section  and  the  mean  length. 
Denoting  this  volume  by  V, 


236.  When  the  cross  section  of  the  ring  is  a  rectangle  the 
volume  may  be  obtained  by  multiplying  the  area  of  a  cross 
section  by  the  mean  length  which  is  the  circumference  of 
a  circle  whose  diameter  is  one  half  the  sum  of  the  inside 
and  outside  diameters  of  the  ring. 


MENSURATION  235 

PROBLEMS 

Find  the  volumes   and  surfaces  of   spheres  having  the 
following  diameters : 

I.  7".       2.  4|".       3.  8'  6".       4.  10.2'.       5.  8000  miles. 

6.  Find  the  weight  of  a  sphere  of  cast  iron  5"  in  diameter. 
(1  cu.  in.  of  cast  iron  weighs  .26  Ib.) 

7.  How  many  cubic  inches  of  water  will  a  hemispherical 
bowl  contain  whose  diameter  is  15"  ? 

8.  What  is  the  weight  of  a  sphere  of  steel  9"  in  diameter, 
if  1  cu.  ft.  of  water  weighs  1000  oz.  and  the  specific  gravity 
of  steel  is  7.8. 

9.  Show  that  the  area  of  the  curved  surface  of  a  hemi- 
sphere is  twice  that  of  its  plane  surface. 

10.  Find  the  weight  of  a  spherical  shell  of  cast  iron 
whose  external  diameter  is  8"  and  whose  thickness  is  ^". 

II.  What  must  be  the  diameter  of  a  hemispherical  basin 
to  hold  1  gal.  of  water  ? 

12.  How  many  yards  of  material  3.25  ft.  'wide  are  neces- 
sary to  make  a  spherical  balloon  containing  1500  cu.  ft. 
of  gas  ? 

13.  Show  that  a  sphere  is  equal  to  two  thirds  of  a  cylin- 
der of  the  same  diameter  and  height. 

14.  A  copper  clothes  boiler  has  semicircular  ends.    Its 
extreme  length  is  20",  its  width  13",  and  its  depth  10". 
If  we  allow  1^-"  on  the  width  of  the  side  piece  for  lock  and 
top  roll,  1"  on  the  length  for  the  side  lock,  and  ^"  all 
around  the  bottom  piece  for  the  lock,  how  much  copper 
is  required  to  construct  the  boiler  ? 

15.  Find  the  volume  of  a  spherical  segment  where  r  =  10", 
r'  =  4",  and  h  =  6". 


236  SHOP  PROBLEMS  IN  MATHEMATICS 

16.  A  bowl  is  in  the  form  of  a  spherical  segment  of  one 
base,  with  r  =  14"  and  k  =  5".    How  many  gallons  will  it 
contain  ? 

17.  How  many  square  feet  of  surface  in  a  zone  whose 
altitude,  h,  is  12',  on  a  sphere  whose  radius  is  21'? 

18.  Find  the  area  and  volume  of  a  ring,  with  a  circular 
cross  section,  whose  mean  radius  is  11  in.  and  whose  in- 
ternal radius  is  10  in. 

19.  The  cross  section  of  a  solid  ring  of  wrought  iron  is  a 
circle  of  6"  radius  and  the  inner  radius  of  the  ring  is  4'. 
Find  (1)  the  area  of  the  surface;  (2)  the  volume  of  the 
ring;    (3)  its  weight  if   1  cu.  in.  of  wrought  iron  weighs 
.27  Ib. 

20.  The  cross  section  of  the  rim  of  a  cast  iron  fly  wheel 
is  a  rectangle  measuring  4  x  8  in. ;  the  inner  diameter  of 
the  rim  is  4'.    Find  (1)  the  area ;   (2)  the  volume ;  (3)  the 
weight  of  the  rim. 


CHAPTER   XVII 

FORMULAS 

In  this  chapter  sufficient  work  in  algebra  is  given  to 
enable  a  student  to  evaluate  the  common  formulas  used 
in  the  shop. 

237.  Law  of  precedence  of  signs.  When  the  signs  -f,  —  , 
X,  and  H-  occur  in  the  same  formula,  the  operations  of 
multiplication  and  division  must  be  performed  before  those 
of  addition  and  subtraction. 

EXAMPLE.    Find  the  value  of  8  x  2  —  4  -;-  2. 


238.  Letters  as  numbers.    The  use  of  letters  for  numbers 
has  been  illustrated  in  sections  18,  47,  60,  in  Chapter  XVI, 
and  in  other  parts  of  the  book.    By  referring  to  section  18 
the  student  will  observe  how  the  same  letter  may  be  used 
to  represent  a  certain  number  in  one  problem  and  another 
number  in  another  problem.    Thus  in  section  18,  problem  1, 
I  —  14,  and  in  problem  2,  I  =  12. 

239.  Factors.    The  factors  of  a  number  are  the  numbers 
that  when  multiplied  together  produce  the  number. 

Thus  2,  3,  and  5  are  the  factors  of  30. 

240.  Three  methods  of  indicating  multiplication.    In  using 
letters  for  numbers  the  product  of  a  and  b  may  be  indicated 
by  any  one  of  the  following  methods  : 

(1)  a  x  b. 

(2)  a  •  b. 

(3)  ab. 
237 


288  SHOP  PROBLEMS  IN  MATHEMATICS 

241.  Terms.    A  term  is  a  number  whose  parts  are  not 
separated  by  a  plus  or  minus  sign ; 

e.g.  6  ab  is  a  term. 

a  -f  b  is  a  number  of  two  terms. 

a  -f  b  —  c  is  a  number  of  three  terms. 

242.  Exponent.    The  exponent  of  a  number  is  a  number 
placed  at  the  right  and  a  little   above   another   number 
called  the  base,  and  indicates  how  many  times  the   base 
is  to  be  taken  as  a  factor. 

Thus  52  =  5  x  5  =  25. 

a8  =  a  X  a  X  a. 

243.  Coefficient.    Any  factor  of  a  term  is  a  coefficient  of 
the  rest  of  the  term  ; 

e.g.  in  4  a,  4  is  the  coefficient  of  a ; 

in  ab,  a  is  the  coefficient  of  b,  and  b  is  the  coefficient 
of  a; 

d  +  d'  f 

in  TT  ?  TT  is  the  coefficient  of 


Unless  otherwise  specified,  the  numerical  coefficient  is  called  the 
coefficient,  as  in  4  a. 

244.  Rule  for  evaluating  an  algebraic  expression. 

(1)  Substitute  for  every  letter  its  numerical  value. 

(2)  Find  the  value  of  every  term  separately,  performing 
the  multiplication  and  division  that  is  indicated. 

This  follows  the  law  of  precedence  of  signs  in  section  237. 

(3)  Combine  the  terms  by  performing  the  addition  and 
subtraction  that  is  indicated. 

(4)  If  a  parenthesis  occurs  in  the  expression,   its  nu- 
merical value  must  be  found  before  combining  with  num- 
bers outside  this  parenthesis. 

The  meaning  of  the  formulas  used  in  the  following  examples  may 
be  found  by  referring  to  the  sections  indicated  at  the  right. 


FORMULAS  239 

Iwt 
EXAMPLE  1.   Evaluate  b  =  —  (Sect.  18) 

when  I  =  15',  w  =  1",  and  t  =  2". 

15-7-2 
~I2~ 
=  17*. 

EXAMPLE  2.   Evaluate  h  =  2  d  +  a  (Sect.  48) 

if  d  =  55'  and  a  =  6'. 

h  =  2  x  55  +  6 
=  110  +  6 


EXAMPLE  3.  Evaluate  F=^(7rr2+7rr'2)+^     (Sect.  229) 
if  h  =  4",  r  =  8",  r'  =  6",  and  TT  =  272. 


32 


=  4_4_Q  0- 
=  J. 


=  662^-  cu.  in. 
(b)  This  problem  may  be  worked  as  follows 


32 


200.+%^      3 

632  TT 
3 

s.  a  . 


cu.  n, 


240  SHOP  PROBLEMS  IN  MATHEMATICS 


ORAL  EXERCISE 

Evaluate  the  following  formulas  with  the  values  indi- 
cated : 

1.  A  =  ab,  if  a  =  10",  b  =  3".  (Sect.  188) 

2.  h  =  ~,  if  d  =  40',  h'  =  5',  d1  =  4',  (Sect.  49) 


b-a,tfl  =  10",  a  =  5".  (Sect.  190) 

4.  A  =  a  (Mjr^V  if  «  =  2",  ft  -  5",  ft'  -  7".   (Sect.  193) 

5.  C  =  Trd,  if  TT  =  -272-,  d  =  21".  (Sect.  198) 

6.  .4  =  ?rr2,  if  TT  =  -272,  r  =  7".  (Sect.  199) 

7.  F  =  aftc,  if  a  =  3",  ft  -  5",  c  =  4".  (Sect.  205) 

8.  d=  Va2  +  ft2  +  c2,  if  a  =  3",  ft  =  4",  c  =  VTi". 

(Sect.  205,  Problem  6) 

9.  V  =  ^B.h,\iB=  12",  h  =  4".  (Sect.  220) 

10.  5  =  4  Trr8,  if  TT  =  -272,  r  =  7".  (Sect.  227) 

11.  F  =  ^-^.7rAr,  if  d=3f,  «*!  =  !',  ^=6.       (Sect.  65) 

12.  ^  =  ?rr2A,  if  r  =  4",  7^  =  14".  (Sect.  213) 

WRITTEN  EXERCISE 

Evaluate  the  following  formulas  with  the  values  indi- 
cated : 

1.  5  =  |  gt*,  ifg  =  32.15,  t  =  4.5.  (Sect.  54) 

2.  c2  =  a2  +  ft2,  if  a  =  7",  ft  =  9".  (Sect.  44) 

3.  a2  =  c2  -  ft2,  if  c  =  10",  ft  =  8".  (Sect.  46) 

4.  7)2  =  ^  +  d^  ifd==  6»  rfi  =  7^  ^gect  45) 

5.  d*  =  D2-  d2,  if  D  =  12",  ^  =  9".  (Sect.  46) 

6.  P  =  Awh,  \i  A  =  125,  w  =  .25,  A  =  10.5".    (Sect.  75) 


FORMULAS  241 


rj      8  =  ¥L,  if  5  =  40,  n  =  25,  TV  =  75.  (Sect.  95) 
?i 

8.  F  =  -  »  if  D  =  3",  JV  =  300.  (Sect.  99) 

9.  l  =  ^-.7rd,ifN=  22°,  d  =  12".  (Sect.  200) 


10.  A=1j--  Trr2,  if  N  =  32°,  r  -  15".  (Sect.  201) 

ooU 

11.  C  =  c  X  E.P.M.,  if  c  =  18",  R.P.M.  =  300.    (Sect.  60) 

12.  I  =  D~tD'  •  TT  +  2rf,  if  D  =  3.5',  Df  =  2.5',  d  =  15'. 

.      (Sect.  62) 


13.    F  =  -  •  TTN,  if  cZ-24",  d1==26",  JV=6.  (Sect.  65) 


14.  H.P.  =         ?r  ?  if  P=8°  lk'  l  =  ^">  A 

-  300.  (Sect.  154,  problem  10) 


15.  A  =  -Vs(s  —  a)(s  —  b)(s  —  c),  if  a  =  24  rd.,  ft  =  30  rd., 
c  =  36  rd.  (Sect.  192) 

16.  ,4  =TT(R*-  r2),  if  ,R  =  16",  r  =  12".  (Sect.  204) 

17.  A=$(p+pl)8,tip  =  20">pl=15",8  =  &'.    (Sect.211) 


18.   V  =        ,  if  d  =  7".  (Sect.  226) 


19.  G=  l'  0034  I,  if  ^  =  20",  d1  =  22",  1=30". 

(Sect.  69) 

20.  V=ih(b  +  b'  +  V^7),  if  A  =  5",  6  =  7",  6'  =  9". 

(Sect.  210) 

245.  Transposition  of  terms.  If  a  line  12"  long  were 
divided  into  two  parts  8"  and  4"  long  respectively,  the 
length  of  the  line  could  be  expressed  by  8"  +  4". 

Evidently  (1)  12"  =  8"  -f  4". 

(2)  12"  -8"  =  4". 


242  SHOP  PROBLEMS  IN  MATHEMATICS 

If  the  length  of  the  line  were  a  and  the  two  parts  re- 
spectively b  and  c,  instead  of  (1)  we  should  have 

(3)  a  =  b  +  c, 
and  instead  of  (2)  we  should  have 

(4)  a-b  =  z. 

Observe  that  b  has  been  changed  from  the  right-hand 
side  of  the  equation  to  the  left-hand  side  by  changing  its 
sign  from  +  to  — . 

From  an  inspection  of  (3)  and  (4),  evidently  —  b  could 
be  changed  back  to  the  right-hand  side  of  the  equation  by 
changing  the  sign  from  —  to  -f- . 

This  changing  from  one  side  of  the  equation  to  the 
other  is  called  transposing. 

Hence  a  term  may  be  transposed  from  one  side  of  the 
equation  to  the  other  by  changing  its  sign. 

EXAMPLE  1.  Solve  the  formula  a  -\-  b  —  c  for  a,  i.e.  find 
the  value  of  a  in  terms  b  and  c. 

(1)  a  +  b  =  c. 

Transposing  b  to  the  right-hand  side  of  the  equation  by 
changing  its  sign,  we  have 

(2)  a  =  c  —  b. 

EXAMPLE  2.    Solve  the  formula  2  s  =  a  -f-  b  +  c  for  b. 

When  no  sign  is  given  to  a  term,  as  in  the  case  of  2  s  in  this 
example,  the  sign  is  understood  to  be  + . 

(1)  2  5  =  a  +  b  +  c. 
Transposing  a  and  c, 

(2)  2  s  -  a  -  c  =  b, 
which  is  evidently  the  same  as 

(3)  I  =  2  s  -  a  -  c. 


FORMULAS 
WRITTEN    EXERCISE 


243 


Given  the  formula 

Solve 
for 

For  meaning  of  formula 
see  section 

1. 

h=  2d+a 

2d 

48 

2. 

k=2d+a 

a 

48 

3. 

c2  =  a2  +  V 

a2 

44 

4. 

S2  =  S2  +  S2 

s2 

44 

5. 

p  =  2a  +  2b 

2  a 

189 

6. 

2m*=b+b' 

b 

193,  note 

7. 

p  =  a  -f  b  +  c 

a 

192,  prob.  5 

8. 

d2  =  D2-  d2 

D2 

46 

9. 

d2  =  D2  -  d2 

4 

46 

10. 

,       D  +  D'    22 

2d 

62 

2       '  7 

246.  How  to  find  the  formula  for  6,  if  ab  =  c. 

If  the  cost  of  3  apples    ==  6^, 
then  the  cost  of  an  apple    =  §/  =  2/. 

Kepresenting  the  number  of  cents  in  the  cost  of  the 
apple  by  a, 


In  the  same  way, 


a  =  §  =  2. 

ab  =  c, 
c 


if 


In  this  equation  a  is  the  coefficient  of  b,  and  the  solution 
consists  in  dividing  both  sides  of  the  equation  by  the  co- 
efficient of  the  unknown  number. 

In  a  similar  way,  dividing  ab  =  c  by  b,  we  have  a  =  -• 


244 


SHOP  PROBLEMS  IN  MATHEMATICS 


EXAMPLE   1.    If    A  =  ab  (sect.   188),   find   the  formula 

for  a. 

^ 

Dividing  by  b,  —  =  a, 

A 

or  a  =  —  • 

EXAMPLE  2.    If  DS  =  ds  (sect.  58),  solve  for  S. 
Dividing  by  D,  S  ==  —  • 

EXAMPLE  3.  The  formula  for  the  area  of  an  ellipse  is 
A  =  Trab  where  a  and  b  are  the  semiaxes.  Solve  the  formula 
for  b. 

A  —  Trab. 

Dividing  by  Tra,  —  =  b, 

Tra 

or  b  = 


WRITTEN    EXERCISE 


Given  the  formula 

Solve 
for 

For  meaning  of  formula 
see  section 

1. 

A  =  ab 

a 

188 

2. 

P  =  br 

b 

167 

3. 

P  =  br 

r 

167 

4. 

A  =2  Trrh 

h 

214 

5. 

V  =  Trr^a 

a 

213 

6. 

L  =  jrda 

d 

214 

7. 

V=abc 

c 

205 

8. 

A  =  ma 

a 

193,  Prob.  2 

9. 

N=PdD 

Pa 

139 

10. 

SNN-iNz  =  sun-in* 

S 

95,  Prob.  2 

FORMULAS  245 

247.  Clearing  of  fractions.  If  we  wished  to  make  f  a 
whole  number,  we  could  do  it  by  multiplying  by  3.  The 
operation  would  be  as  follows  : 

2 

v  o  _  o 

? 

In  the  same  way  we  may  make  the  following  fractions 
into  whole  numbers  by  multiplying  by  the  number  indicated: 


If  we  wish  to  change  the  formula 

A  =  y  (Sect.  190) 

so  that  there  shall  be  no  fraction  in  it,  we  may  multiply 
both  sides  of  the  equation  by  2  and  obtain 

2^=yx?, 
or  2  A  =  bk. 

The  advantage  of  this  operation  is  now  apparent,  since 
we  may  solve  for  b  or  li  by  section  246. 

EXAMPLE  1.    In  the  formula  V  =  —  (sect.  206)  clear  of 
fractions  and  solve  for  h. 

VJ*L 

~  3' 

Multiplying  by  3,  3  V  =  bk. 

3  V 

Dividing  by  b}  ——  =  h, 

3V 

or  h  =  —  • 

b 


246  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  2.    Solve  the  formula  Pc  —  ——  (sect.  146)  for  N. 

P  =  ^. 

c        N 

Multiplying  by  N,        NPC  =  TrD. 
Dividing  by  Pc,  N  =  -  — 

EXAMPLE  3.    Solve  I  =  TT  D  "t  D-  +  2d  (sect.  62)  for  d. 


. 

Multiplying  by  2,  21  =  TT(D  +  D1)  +  4  d. 

Transposing,  2  I  -  TT(D  +  D1)  =  4  d. 

Dividing  by  4,        2* 


EXAMPLE  4.    Solve  —  =  —  for  TV  (sect.  95). 
*       N 

In  this  case  we  clear  of  fractions  by  multiplying  by  the 
product  of  the  two  denominators. 

S  n 

—  x  sN  =  —  X  sN. 

s  N 

SN  =  ns. 

7?*? 

Dividing  by  5,  N  =  — 

S 

248.  When  the  unknown  number  has  an  exponent  2  or  3. 

EXAMPLE  1.    Solve  A  =  a2  for  a  (sect.  188). 

.1  =  a2. 
Taking  the  square  root  of  both  members  of  the  equation, 


or 


FORMULAS  247 

EXAMPLE  2.    Solve  V  =  as  for  a  (sect.  205,  Problem  4). 

V=as. 
Taking  the  cube  root  of  both  members, 


or  a  = 

The  student  should  interpret  the  results  of  all  transformed  formulas. 

EXAMPLE  3.    Solve  V  =  7rr2a  for  r  (sect.  213). 
F  =  7rr*a. 

Dividing  by  Tra,  -  =  r2, 

or  r2  = 

Tra 

Taking  the  square  root  of  both  members  of  the  equation, 


r  = 


EXAMPLE  4.    Solve  V=7J^~  for  d  (sect.  226). 


_ 

:    6 


Multiplying  by  6,        6  V  = 
Dividing  by  TT,  -  =  d3, 

6V 

or  d*  =  — 

7T 

(or 

Taking  cube  root,  d  =  Jj  --- 

^    7T 

In  the  following  exercise  the  answers  may  be  checked  by 
reversing  the  operations  and  solving  the  resulting  formula 
for  the  letter  that  was  first  given.  The  student  should 
interpret  the  meaning  of  the  given  formula  and  the  result- 
ing formula. 


248 


SHOP  PROBLEMS  IN  MATHEMATICS 
WRITTEN   EXERCISE 


Given  the  formula 

Solve 
for 

For  meaning  of  formula 
see  section 

1. 

t_Pc 

PC 

143 

2 

2. 

A  -  kp 

A 

197 

2 

3. 

h      dh/ 

K 

49 

~~¥ 

ri                   1  * 

j. 

. 

K 

2 

n 

AT" 

«K 

o. 

2 

a.V 

oo 

6. 

P  _w 

10 

177 

W~  p 

P  _w 

7. 

W~p 

P 

177 

8. 

P  _w 

W 

177 

9. 

s=  16.  08  *2 

t 

54 

10. 

c2  =  a2  +  &2 

a 

188,  Prob.  7 

11. 

-4  —  — 

d 

199 

4 

12. 

I  _   N  .nd 

N 

200 

360  ' 

13. 

S  =  47ZT2 

r 

227 

14. 

5  -  7T#> 

d 

227 

15. 

OD-D+   2 

Pd 

138 

Pd 

16. 

JV" 

r 

201 

~  360'7 

17. 

F=27T2r2# 

R 

235 

18. 

T7-  _          2      ^'  +  h" 

T 

217 

2 

19. 

^1  =  ?r(jR2  -  r2) 

R 

204 

4 

20. 

F  =  -  TIT3 

r 

226 

3 

CHAPTER   XVIII 

SOLUTION  OF  RIGHT  TRIANGLES  BY  NATURAL  FUNCTIONS 

249.  The  trigonometric  functions.  If  B  is  a  point  on  the 
side  of  the  Z.  EAD  and  BC  is  drawn  perpendicular  to  AD 
at  C,  forming  the  right  triangle  ABC, 

-  is  called  the  sine  of  the  angle  A. 
c 

-  is  called  the  cosine  of  the  angle  A. 

G 

-  is  called  the  tangent  of  the  angle  A. 

-  is  called  the  cotangent  of  the  angle  A. 

Ct 

-  is  called  the  secant  of  the  angle  A. 

C  • 

-  is  called  the  cosecant  of  the  angle*^4.    * 

The  sine,  cosine,  tangent,  etc.,  are  called  trigonometric 
functions  of  the  angle  A .  &e. 

The  value  of  each  of  these  ratios 
may  be  found  approximately  by 
measuring  a,  b,  and  c,  and  rinding 

the  numerical  value  of  the  ratios.    A'  b       C        c'  ~ 

If  B  is  taken  at  any  other  point  FlG    155 

on  A  E,  such  as  B',  it  can  be  shown 
by  measurement  or  by  similar  triangles  that  the  ratios  ->  -? 

etc.,  will  have  the  same  values  as  before,  though  a,  b,  and  c 
will  vary. 

249 


250 


SHOP  PROBLEMS  IN  MATHEMATICS 


i 


EC        1 

In  Fig.  156,  sine  A  =  -.-  =  ±  =  .5. 
AJ3       A 

Since  the  sine  does  not  vary  for  a  given  angle,  if  the 
angle  is  given,  the  sine  is  determined. 

The  sine,  cosine,  tangent,  co- 
,B'     tangent,  seoant,  and  cosecant 
are  called   natural    functions 
of  the  angle,  since  for  every 

J  change  in  the  angle  there  is 
a  change  in  the  value  of  the 
sine,  cosine,  etc. 

The  table  on  page  279  gives 
the  values  of  these  functions  for  angles  from  0°  to  90°. 

In  machine  work,  engineering,  and  surveying  it  is  neces- 
sary to  know  how  to  use  these  functions  in  the  solution  of 
triangles. 

EXAMPLE  1.    Given  A  =  35°  30',  c  =  32 ;  to  find  B,  a,  1>. 


B  =  90°  -  35°  30'  =  54°  30'. 

a 

c. 


a  —  c  sine  A. 


From  the  tables, 

sin  A  =      .5807 
c  =  32 

11614 
17421 


-  =  cos  A. 
c 


cos  A  =      .8141 
c=  32 


a  =  18.5824 
=  18.6 


16282 
24423 


I  =  26.0512 


Check. 


322  = 


1024  =  346  +  678 
1024  =  1024 


SOLUTION  OF  RIGHT  TRIANGLES  251 

EXAMPLE  2.    Given  A  and  a ;  to  find  B,  c,  and  b. 
£  =  90°  -  A. 

a  a 

—  =  sin  A  ;  c  = 


c  sin.  A 

-  =  cos  A  :  b  =  c  cos  A. 
c 

Check  as  in  Example  1. 

EXAMPLE  3.    Given  A  and  b ;  to  find  B,  c,  and  a. 

B  =  90°  -  A. 

b  b 


-  =  cos  A  :    c  — 

c  cos  A 

a 

—  =  sin  A  ;  a  =  c  sin  A. 

Check  as  before. 

EXAMPLE  4.    Given  a  and  c;  to  find  A,  B,  and  &. 

a 

sin  A  =  -• 
c 

&  =  c  sin  ^4 . 
Check  as  before. 

EXAMPLE  5.    Given  a  and  b ;  to  find  A,  B,  and  c. 

a 
tan  4  =  —  • 

B  =  90°  -  ,4. 

a 
c  =  — 

Check  as  before. 

WRITTEN   EXERCISE 

To  solve  a  triangle  means  to  find  the  unknown  sides  and  angles. 
Solve  and  check  the  right  triangles  given : 

1.  ,4=42°,  c  =  92.4. 

2.  A  =  25°,  a  =  56.1. 


252  SHOP  PROBLEMS  IN  MATHEMATICS 

3.  A  =50°,  5  =  17.3. 

4.  a  —  75,  c  =  85.5.  (Find  angles  correct  to  degrees.) 

5.  a  =  82,  5  =  64.      (Find  angles  correct  to  degrees.) 

6.  £  =  74°  30',  a  =  142.3. 

To  find  the  sine  of  74°  30' ;  to  the  sine  of  74°  add  half  the  difference 
between  the  sine  of  74°  and  75°.  Thus  the  sine  of  74°  is  .9613,  and  the 
sine  of  75°  is  .9659.  One  half  the  difference  is  .0023,  which,  added  to 
.9613,  gives  .9636  for  the  sine  of  74°  30". 

Similarly,  for  an  angle  of  15' use  one  fourth  of  the  difference,  and  for 
45'  use  three  fourths. 

7.  B  =  15°  15',  c=24.6. 

8.  yl  =  64°45',  5  =  18.4. 

9.  a  =  16.2,  5  =  22.5.    (Find  angles  correct  to  15'.) 

10.  £  =  83°  30',  5  =  125.7. 

11.  A  disk  is  10.5"  in  diameter.    Find  the  distance  neces- 
sary to  set  a  pair  of  dividers  in  order  to  space  off  5  sides 
(i.e.  to  inscribe  a  regular  polygon  of  5  sides  in  the  disk). 

12.  Find  the  number  of  gallons  of  oil  in  a  cylindrical 
tank  20'  long,  whose  head  has  a  diameter  of  6',  if  a  pole 
thrust  through  a  hole  in  the  top  of  the  tank  to  the  bottom 
is  found,  on  being  pulled  out,  to  be  covered  with  oil  to  the 
length  of  2'  4". 

13.  From  the  preceding  problem  derive  a  formula  for  the 
volume  of  oil  in  a  cylindrical  tank  in  terms  of  d,  the  diam- 
eter of  the  head  of  the  tank ;  Z,  its  length  ;  and  s,  the  part  of 
the  radius  between  the  chord  and  the  arc,  i.e.  the  length 
of  the  pole  that  was  wet  with  oil. 

14.  Derive  a  formula   for   the  exact  length  of   a   belt 
around  two  pulleys  in  each  of  the  following  cases : 

(a)  Open  belt  on  pulleys  of  diameters  D  and  d,  the  dis- 
tance between  centers  being  a  (see  Fig.  38,  p.  70). 

(5)  Cross  belt  on  the  same  pulleys  (see  Fig.  40,  p.  71). 

For  other  problems  involving  trigonometry,  see  sections  147-149. 


CHAPTER   XIX 

SHORT  METHODS 

250.  Accuracy  and  speed.    Most  of  the  mathematical  oper- 
ations that  are  used  in  the  trades  and  in  business  are  very 
simple,  but  in  their  use  accuracy  is  absolutely  essential  and 
speed  is  desirable.    Accuracy  may  be  secured  by  practice 
and  by  checking  all  work. 

Speed  may  be  improved  by  (1)  a  knowledge  of  short 
methods,  (2)  by  noting  the  time  required  to  perform  the 
operation,  and  repeating  the  work,  trying  to  reduce  the 
time. 

251.  Addition.    (1)  When  adding   a  column,  try   to   see 
combinations  that  make  10,  such  as  6  and  4,  7  and  3,  or 
5,  4,  and  1. 

(2)  When  numbers  do  not  combine  readily  into  10,  try 
for  9  or  11. 

(3)  Practice  adding  columns  that  add  to  more  than  199. 

(4)  Check  by  adding  in  reverse  order. 

(5)  Learn  to  detect  errors  by  dividing  the  column  into 
several  parts  and  adding  each  separately. 

(6)  Time  yourself  when  doing  a  problem.    Repeat  the 
addition  with  the  same  or  a  similar  problem  and  try  to 
reduce  the  time. 

(7)  An  error  in  copying,  due  to  transposition  of  figures, 
is  divisible  by  D.    Thus  75  copied  57  makes  an  error  of  18, 
which  is  divisible  by  9. 

The  proof  of  this  is  as  follows  for  a  number  of  two  digits  : 
(10 a  +  6)-(o+  106)  =  9a-96  =  9(a  -6). 
253 


254  SHOP  PROBLEMS  IN  MATHEMATICS 

WRITTEN   EXERCISE 


h 

k 

I 

m 

a 

3478.92 

9136.07 

721.3 

4278.91 

b 

6782.75 

349.25 

842.65 

542.31 

c 

702.43 

7865.50 

384.99 

465.78 

d 

685.21 

2842.65 

371.46 

1942.85 

e 

4889.75 

5600. 

768.24 

1392.99 

f 

4672.44 

8929.48 

3421.18 

9276.49 

9 

3498.25 

6854.35 

8421.55 

3498.46 

Write  in  a  column  each  combination  of  rows  or  columns 
indicated  below,  add,  and  check  by  adding  in  reverse  order. 

I.  a  +  b  +  c.  6.  k  +  I. 


2.  d  +  e+f+g. 

3.  a  -f-  c  -f-  &  +  g. 

4.  i  4.  d  +  /. 
5. 


7.  k  +  m. 

8.  h  +  k  -f  I. 

9.  k  +  I  +  m. 
10.  A  +  k  +  Z 


252.  Multiplication.    (1)  To  multiply  by  10  move  the  deci- 
mal point  one  place  to  the  right  ;  e.g.  25.67  x  10  =  256.7. 

(2)  To  multiply  by   100  move  the  decimal  point  two 
places  to  the  right  ;  e.g.  2.5673  x  100  =  256.73. 

(3)  To  multiply  by  5  (V)  divide  by  2  and  multiply  by  10  ; 

e.g.  3842  x  5  =  S^SL  x  10  =  1921  x  10  =  19,210. 

(4)  To  multiply  by  3£  (y)  divide  by  3  and  multiply 
by  10  ;  e.g.  5670  x  3J  =  Mp  X  10  =  1890  x  10  =  18,900. 

(5)  To  multiply  by  2^  (-L°-)  divide  by  4  and  multiply 
by  10  ;  e.g.  16  x  2£  =  ^  x  10  =  4  x  10  =  40. 

(6)  To  multiply  by  50  (J-§£)  divide  by  2  and  multiply 
by  100  ;  e.g.  42  x  50  =  -\2-  x  100  =  21  x  100  =  2100. 


SHORT  METHODS  255 

(7)  To  multiply  by  25  (L$&).  divide  by  4  and  multiply 
by  100 ;  e.g.  48  x  25  =  V  X  100  =  12  x  100  =  1200. 

(8)  To  multiply  by  12|  (i§&)  divide  by  8  and  multiply 
by  100 ;  e.g.  24  x  12J  =  -2/-  X  100  =  3  x  100  =  300. 

(9)  To  multiply  by  33J  (ijp)  divide  by  3  and  multiply 
by  100 ;  e.g.  27  x  33J  =  V  x  100  -  9  x  100  =  900. 

(10)  To  multiply  by  16§  (L$a)  divide  by  6  and  multiply 
by  100  ;  e.g.  42  x  16$  -  V  X  100  -  7  x  100  =  700. 

(11)  To  multiply  by  14f  (1^)  divide  by  7  and  multiply 
by  100  ;  e.g.  35  x  14f  =  *f  x  100  =  5  x  100  =  500. 

(12)  To  multiply  by  66|  (f  of  100)  multiply  by  100  and 
subtract  £  of  the  product  from  it ; 

e.g.  39  x  66f  =  39  x  100  -  a ^QO  =  2600. 

(13)  To  multiply  by  333^  (1%°-*)  divide  by  3  and  multiply 
by  1000;  e.g.  21  x  333 J  -  V  X  1000  =  7000.. 

(14)  To  multiply  by  166|  (1%°--)  divide  by  6  and  multi- 
ply by  1000  ;  e.g.  54  x  166§  =  -%*  x  1000  =  9000. 

(15)  To  multiply  by  125  (^%°-^-)  divide  by  8  and  multiply 
by  1000  ;  e.g.  96  x  125  =  *£  x  1000  =  12,000. 

(16)  To  multiply  by  9  (10  —  1)  multiply  by  10  and  sub- 
tract the  multiplicand  from  the  product ; 

e.g.  23  x  9  =  23  x  10  -  23  =  207. 

(17)  To  multiply  by  11  (10  + 1)  multiply  by  10  and  add 
the  multiplicand  to  the  product ; 

e.g.  34  x  11  =  34  x  10  +  34  =  374. 

(18)  When  one  number  is  even  and  is  between  10  and 
20  multiply  half  of  this  number  by  double  the  other. 

e.g.  37  x  18  =  54  x  9  =  486. 


256  SHOP  PROBLEMS  IN  MATHEMATICS 

(19)  To  square  a  number  of  two  figures  with  5  in  units 
place  multiply  the  tens'  figure  by  one  more  than  itself  and 
annex  25  to  the  product ; 

e.g.  652  =  6  x  7  with  25  annexed  =  4225. 
The  reason  for  this  may  be  seen  from  the  algebraic  form  (10a  +  5)2. 
(10  a  +  5)2  =  100  a2  + 100  a  +  25  -  100  a  (a  +  1)  +  25. 

(20)  To  square  a  mixed  number  with  1  for  the  fractional 
part,  multiply  the  whole  number  by  one  more  than  itself 
and  add  J  ;  e.g.  7i2  =  7  x  8  +  J  =  66J. 

This  may  be  proved  as  follows  : 

(a  +  i)2  =  a2  +  a  +  \  -  a  ( a  +  1)  +  \. 

(21)  When  the  numbers  differ  by  one,  to  the  square  of  the 
smaller  add  the  smaller  ;  e.g.  15  x  16  =  152  +  15  =  240. 

(22)  To  multiply  by  a  number  that  is  one  more  or  less 
than  a  multiple  of  10,  proceed  as  in  the  following  examples  : 

EXAMPLE  1.  28  x  19  =  28  x  20  -  28  =  560  -  28  =  532. 
EXAMPLE  2.  28  x  21  =  28  x  20  +  28  =  560  +  28  =  588. 

(23)  To  multiply  by  15,  multiply  by  10  and  add  to  the 
product  £  of  itself  ;  e.g.  48  x  15  =  480  -f  240  =  720. 

ORAL    EXERCISE 


1. 

3.457x10. 

10. 

126  x  16}. 

19. 

(35)*. 

2. 

23.6  x  100. 

11. 

357  x  14f  . 

20. 

(9i)2- 

3. 

7234  x  5. 

12. 

27  x  66|. 

21. 

84  x  25. 

4. 

234  x  3f 

13. 

48  x  333^. 

22. 

34  x  19. 

5. 

92  x  2|. 

14. 

42  x  166}. 

23. 

(75)2. 

6. 

7893  x  50. 

15. 

56  x  125. 

24. 

(5i)2- 

7. 

76  x  25. 

16. 

35x9. 

25. 

(15i)2. 

8. 

272  x  12|. 

17. 

42  x  11. 

26. 

16x17. 

9. 

126  x  33i. 

18. 

13  x  14. 

27. 

34  x  15. 

SHORT   METHODS  257 


28. 

53x 

9. 

35. 

78 

X 

9. 

42. 

(!' 

H 

)2. 

29. 

674  x  5. 

36. 

54 

X 

66J. 

43. 

27 

x 

15. 

30. 

639  x  3i. 

37. 

84 

x 

33J. 

44. 

32 

X 

11. 

31. 

76  x 

2f 

38. 

96 

X 

333J. 

45. 

16 

X 

19. 

32. 

92  x 

25. 

39. 

64 

X 

125. 

46. 

27 

x 

33£. 

33. 

66  x 

16f. 

40. 

452. 

47. 

72 

X 

25. 

34. 

504  x  14f  . 

41. 

m 

ff 

! 

48. 

66 

X 

66|. 

253.  Division.    (1)   To  divide  by  10  move  the  decimal 
point  one  place  to  the  left; 

e.g.  34.5  -f-  10  =  3.45. 

(2)  To  divide  by  100  move  the  decimal  point  two  places 
to  the  left  ;  ^  ^3  +  i00  =  2.456. 

(3)  To  divide  by  5  (-V0-)  multiply  by  2  and  divide  by  10  ; 

.  0.,  ...          r         4olO  X    ^         or»o 

e.g.  4315  -j-  5  =  -  —  --  =  863. 

(4)  To  divide  by  3^  (  L°-)  multiply  by  3  and  divide  by  10  ; 

e.g.  231  ^3^  =  ^j^  =  69.3. 

(5)  To  divide  by  2i  (i£)  multiply  by  4  and  divide  by  10  ; 

e.g.  52  +  2i  =  ^  =  20.8: 

(6)  To  divide  by  50  multiply  by  2  and  divide  by  100  ; 
r      '  e.g.  378  +  50  = 


(7)  To  divide  by  33J  (iga)  multiply  by  3  and  divide  by 

100  5  132  xS 

e.g.  132  +  33J  =  =  3.96. 


258  SHOP  PROBLEMS  IN  MATHEMATICS 

(8)  To  divide  by  25  multiply  by  4  and  divide  by  100; 
V-  ^300.25  = 


(9)  To  divide  by  16§  multiply  by  6  and  divide  by  100  ; 

'  e.g.  210.  16§  =  ^_2p  =.12.60. 

(10)  To  divide  by  14f  multiply  by  7  and  divide  by  100  ; 

e.g.  320  -,-l4»=??°_>p  =  22.4o. 

(11)  To  divide  by  12|  multiply  by  8  and  divide  by  100; 

=  32.80. 


(12)  To  divide  by  333i  multiply  by  3  and  divide  by 

2321  x  3 
e.g.  2321  H-  333J  =    **  *  =  6.963. 


(13)  To  divide  by  166§  multiply  by  6  and  divide  by 
e.g.  3200  +  166§  =  ~  =  19.20. 


(14)  To  divide  a  number  by  66§  add  to  the  number 
of  itself  and  divide  by  100  ; 

200  +  100 


e.g.  « 

100 

—  t. 

>. 

ORAL  EXERCISE 

1.  4.532-10. 

6.  427-5-50. 

11. 

30  -  12J. 

2.  3.245-100. 

7.  62  -  33^. 

12. 

2120  -  333J. 

3.  3240  -f-  5. 

8.  2300  -=-  25. 

13. 

700-3-166}. 

4.  322-3i. 

9.  306  -i-  16}. 

14. 

400  -5-  66}. 

5.  33  -2|. 

10.  41-f-14f. 

15. 

72  -  33J. 

SHORT  METHODS  259 

16.  350  -5-  25.  19.  (55)2.  22.  36  x  25. 

17.  200  -f-  16|.  20.  (9£)2.  23.  31  x  11. 

18.  24-i-12£.  21.  19x16.  24.  42x15. 

25.  Multiply  256  by  5.  33.  62£  +  5. 

26.  Multiply  480  by  25.  34.  721£  -5-  25. 

27.  Multiply  320  by  125.  35.  201£  -f- 125. 

28.  Multiply  660  by  33£.  36.  64  x  25. 

29.  Multiply  422  by  5.  37.  201|  -*-  25. 

30.  Multiply  524  by  25.  38.  72  x  125. 

31.  Multiply  136  by  33£.  39.  (125)2. 

32.  Multiply  720  by  125.  40.  (16i)2. 


CHAPTER  XX 

THE  SLIDE  RULE      ., 

254.  Use.  The  slide  rule,  also  called  the  calculating 
rule  and  the  guessing  stick,  is  an  instrument  for  rapid  cal- 
culations in  operations  involving  multiplication,  division, 
squares  and  square  root,  cubes  and  cube  root,  natural  trig- 
onometric functions,  and  logarithms  of  numbers.  Addition 
and  subtraction  cannot  be  performed  on  this  rule. 


•1                   2.         3      A     5    6, 

7  8  9.  10                20 

30     40   50  60  70  GOX  loo 

A 

11    1 

\         1 

11  1 

R                            'I                 'I 

1    1 

1         I 

1  1 

2          3456 
'                                        * 

7  Q  9  10                 20          30     40   JO  60  708030  MW 
3                -4             5         t>         7      &      9     to 

u    f  

1  1— 

1  

—  1  1  

Dl  1  

3             4567 

83/0 

FIG.  158.    SLIDE  RULE 

The  results  are  usually  correct  to  three  significant  figures, 
which  is  sufficiently  accurate  for  ordinary  work. 

255.  Principles  of  its  use. 

102  =  100. 
Another  method  of  stating  this  equation  is 

The  logarithm  of  100  is  2. 
Similarly  108  =  1000. 

Hence  the  logarithm  of  1000  is  3. 
Note  that  the  logarithm  is  the  exponent  which  is  given  to  10. 

256.  Law  of  multiplication. 

100  x  1000  =  100,000, 

and  105  =  100,000, 

or  log  100,000  =  5 

-2  +  3 

=  log  100  +  log  1000. 

260 


THE  SLIDE  RULE 


261 


It  is  a  principle  of  loga- 
rithms, as  may  be  seen  in 
this  example,  that  the  loga- 
rithm of  a  product  is  the  sum 
of  the  logarithms  of  the 
multiplicand  and  multiplier. 
Hence  to  multiply  one  num- 
ber by  another  add  their  log- 
arithms. 

257 .  Description  of  the  slide 
rule.  The  slide  rule  consists 
of  a  grooved  stick  about  10" 
long,  carrying  a  movable 
slide.  There  are  four  scales, 
A,  B,  C,  and  D,  as  in  the  fig- 
ure, divided  proportionally 
to  the  logarithms  of  the  num- 
bers that  are  marked  upon 
them.  Scale  A  is  marked 
exactly  like  B,  and  scale  C 
like  D. 

EXAMPLE  1.  Suppose  we 
wish  to  multiply  2  by  3,  us- 
ing the  slide  rule  with  the 
upper  scales  A  and  B. 

Move  the  slide  to  the  right 
until  1  on  scale  B  is  under 
2  on  scale  A.  Then  over  3 
on  B  find  the  product  6  on  A. 
Here  we  have  added  log  2 
and  log  3,  obtaining  log  6. 

If  we  use  the  lower  scales 
C  and  D,  move  the  slide  to 
the  right  till  1  on  scale  C 


262  SHOP  PROBLEMS  IN  MATHEMATICS 

is  above  2  on  scale  D.   Then  under  3  on  C  find  Jbhe  product 
6  on  D. 

258.  Finding  numbers  on  the  slide  rule. 

ORAL   EXERCISE 

1.  On  scales  A  and  B  find  the  following  numbers,  check- 
ing by  referring  to  Fig.  159  :  1.24,  1.85,'  1.97,  2.62,  9.45, 
17.3,  8.68,  9.25. 

2.  On  scales  C  and  D  find  1.24,  1.85,  2.62,  and  check  by 
referring  to  Fig.  159. 

259.  When  to  use  scales  A  and  B.   If  only  approximate 
results  are  desired  and  the  numbers  involved  have  a  wide 
range  of  values  use  scales  A  and  B. 


•                                       48 
r  1  

35.5 
^1  

\ 

1  !  

1 
7.4 

FIG.   160.    4.8  x  7.4  =  35.5.    SCALES  A  AND  B. 

260.  When  to  use   scales  C  and  D.    If  greater  accuracy 
is  desired,  use  scales  C  and  Z>.    For  shop  work  scales  C  and 
D  are  more  often  used  than  A  and  B. 

261.  Multiplication. 

EXAMPLE  1.    Multiply  4.8  by  7.4  (see  Fig.  160). 

Set  1  on  B  to  4.8  on  A,  and  above  7.4  on  B  read  35.5  on  A. 

3.62 

.  I S=          =" 

2-35  S.5 

FIG.  161.    2.35  x  3.62  =  8.50.    SCALES  C  AND  D 

EXAMPLE  2.    Multiply  2.35  by  3.62  (see  Fig.  161). 

Set  1  on  C  to  2.35  on  D,  and  under  3.62  on  C  read  8.50  on  D. 

EXAMPLE  3.    Multiply  23.5  by  36.2. 

23.5  x  36.2  =  2.35  x  3.62  x  100 

=  8.50  x  100  (by  Example  2) 

=  850. 


THE  SLIDE  RULE  263 

EXAMPLE  4.    Multiply  5.347  by  172.3. 

Taking  each  number  correct  to  three  significant  figures, 
we  multiply  5.35  by  172.  Set  1  on  C  to  5.35  on  D,  and  under 
1.72  on  C  find  9.20  on  D. 

5.35  x  172  =  5.35  x  1.72  x  100 
=  9.20  x  100 
=  920. 

If  the  fourth  significant  figure  is  5  or  greater  than  5,  we  increase 
the  third  figure  by  1.  If  this  occurs  in  both  numbers,  a  more  accurate 
result  will  be  secured  by  increasing  the  third  figure  in  only  one  num- 
ber. Thus  5.435  x  7.018  should  be  worked  as  5.43  x  7.02. 


FIG.  162.    3.41  x  6.85  =  23.4. 

EXAMPLE  5.    Multiply  3.41  by  6.85  (see  Fig.  162). 

Set  1  on  C  to  3.41  on  D,  and  6.85  on  C  projects  beyond 
the  end  of  D.  Set  10  on  C  to  3.41  on  D,  and  under  6.85  on 
C  find  2.34  on  D.  Setting  10  instead  of  1  opposite  the  first 
number  divides  the  result  by  10.  Hence  we  must  multiply 
2.34  by  10.  The  result  is  finally  23.4.  In  such  a  case  the 
position  of  the  decimal  point  may  be  found  by  inspection. 
If  scales  A  and  B  are  used  and  the  slide  projects  to  the  left, 
use  100  as  the  index. 

EXAMPLE  6.    Multiply  .00246  x  .0565. 
Using  scales  C  and  Z>, 

.00246  x  .0565  =  2.46  x 10~3  x  5.65  x  10~2 
=  2.46  x  5.65  x  10- 5 
=  13.9xlO-5 
=  .000139. 

10-3  =  ±:      J_ 

103      1000 


264  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  7.  Another  method  of  obtaining  the  decimal 
point  in  Example  6. 

.00246  x  .0565  is  roughly  .002  x  .06  =  .00012. 

From  this  we  see  that  the  significant  figures  139  should 
be  read  .000139. 

EXAMPLE  8.    Find  the  product  of  4,2  x  65  x  .34  x  2.6. 

Using  scales  C  and  D,  set  10  on  C  to  4.2  on  D  and  bring 
the  runner  to  6.5  on  C. 

Bring  1  on  C  to  the  runner  line,  then  the  runner  to  3.4 
on  C,  10  on  C  to  the  runner.  Under  2.6  on  C  read  the 
result,  241,  on  D. 

The  position  of  the  decimal  point  is  determined  by  a 
rough  calculation. 

4.2  x  65  x  .34  x  2.6  is  roughly  4  x  60  x  .5  x  2  =  240. 

Hence  the  result  is  241. 

ORAL   EXERCISE 

Using  scales  C  and  D,  find  the  value  of : 

1.  2.32  x  3.14.  8.  4.51  x  7.94. 

2.  3.78  x  5.26.  9.  .00382  x  .00475. 

3.  1.79  x  3.25.  10.  9.84  x  .000262. 

4.  8.62  x  3.44.  11.  4673  x  842.5. 

5.  2.3  x  3.16.  12.  .641  x  .032. 

6.  32.5  x  25.2.  13.  8.2  x  32  x  .45  x  6.4. 

7.  6.375  x  281.3.  14.  .75  x  22  x  6.5  x  8.65. 

15.  Find  the  circumferences  of  circles  having  diameters 
of  3',  4.5',  15". 

C  =  TT  D.  Set  1  on  B  to  TT  on  A  and  above  3,  4.5,  and  15 
on  B,  read  the  circumferences  on  A. 

16.  Find  the  lateral  area  of  a  cylinder  whose  length  is 
16f  and  the  diameter  of  whose  base  is  7.5'. 

Lateral  area  •=  length  x  TT  x  diameter  of  base. 


THE  SLIDE  RULE  •  265 

Using  scales  A  and  B,  find  values  of : 

17.  7.82  x  0.48.  19.  13.4  x  8.42. 

18.  11.5  x  8.35.  20.  6.76  x  14.5. 

262.  Division.  Since  division  is  the  reverse  process  of 
multiplication,  it  may  be  performed  on  the  slide  rule  by 
reversing  the  operations  for  multiplication. 

EXAMPLE  1.    Divide  6  by  3. 

Set  3  on  B  under  6  on  A  and  over  1  on  B,  read  the 
quotient  2  on  A . 

In  finding  6  on  A  it  may  be  convenient  to  use  the  runner. 

EXAMPLE  2.    Divide  35.5  by  4.8. 

Set  4.8  011  B  to  35.5  on  A  and  over  1  on  B,  read  the 
quotient  7.4  on  A. 

EXAMPLE  3.    Divide  8.5  by  2.35,  using  scales  C  and  D. 

Set  2.35  on  C  to  8.5  on  D  and  under  1  on  C,  read  the 
quotient  3.62  on  D. 

EXAMPLE  4.    Divide  850  by  23.5. 

This  is  the  same  as  Example  3  except  that  the'  decimal 
point  is  evidently  after  the  6,  making  the  quotient  36.2. 

EXAMPLE  5.    Divide  9.203  by  5.347. 

Using  three  significant  figures,  this  becomes  9.20  -j-  5.35 
=  172. 

EXAMPLE  6.    Divide  2.34  by  3.45. 

Set  3.41  on  C  to  2.34  on  D  and  opposite  10  on  C,  read 
the  quotient  6.85  on  D. 

If  scales  A  and  B  are  used,  over  the  index  100  on  J3,  read 
the  significant  figures  and  place  the  decimal  point  by  in- 
spection. 

EXAMPLE  7.  .000139  -f-.0565=(l. 39  x  10~4) -=-(565  xlO~2) 

=  (1.39-5-  5.65)  x  10- 2 
=  .  246xlO-2  =  .00246. 


266  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  8.    Another  method  of  obtaining  the  decimal 
point  in  Example  7. 

.000139  -f-  .0565  is  roughly  .0001  -r-  .05  =  .002. 
From  this  we  see  that  the  significant  figures  246  should 
be  read  .00246. 

ORAL  EXERCISE 

Using  scales  A  and  B,  find  the  value  of : 

1.  65.5 -f- 3.45.  3.  80.5-^1.75. 

2.  100  -h  39.5.  4.  14.4  --  16.5. 

5.  1.63  -r-  3.1416  (3.1416  =  TT). 
Using  scales  C  and  D,  find  the  value  of : 

6.  5.48  H-  2.44.  13.  1.38  -=-  5.65. 

7.  7.35-3.13.  14.  1670 -.375. 

8.  6.13  -  4.62.  15.  .00665  -  .000144. 

9.  9.65  -  8.45.  16.  .285  -=-  .075. 

10.  10  -  3.1416.  17.  2580  -=-  .187. 

11.  3.64-6.45.  18.  .00156-32.8. 

12.  4.35  -  7.52.  19.  .375  -  .065. 

20.  .0000038537  -  .0011887. 

Reduce  to  decimals,  correct  to  three  significant  figures. 
Divide  the  numerator  by  the  denominator. 

21.  TV  23.    §£.          25.   TV          27.   |f.          29.   }f. 

22.  if.  24.   |if.        26.   Jf.  28.   if^        30.   6jff. 

263.  Reciprocals.   Two  numbers  are  reciprocals  if  their 
product  is  equal  to  1.    Thus  the  reciprocal  of  3  is  -£•;  the 
reciprocal  of  §  is  | ;  the  reciprocal  of  ^  is  3. 

264.  Reciprocals  by  the  inverted  slide.    If  the  slide  be  in- 
verted so  that  the  C  scale  is  in  contact  with  the  A  scale, 
and  the  right-hand  index  (10)  of  the  slide  be  placed  in  co- 
incidence with  the  left-hand  index  (1)  of  the  A  scale,  every 


THE  SLIDE  RULE  267 

number  on  the  D  scale  multiplied  by  one  tenth  of  the  cor- 
responding number  on  the  inverted  C  scale  is  equal  to  1 ; 

e.g.  2  x  .5  =  1. 

Hence,  to  find  the  reciprocal  of  a  number,  invert  the  slide, 
set  10  on  the  slide  to  1  on  scale  A,  and  one  tenth  of  every 
number  on  the  inverted  scale  C  will  be  the  reciprocal  of  the 

corresponding  number  on  scale  D ; 

2.5 
e.g.  the  reciprocal  of  4  is  -3***  =  .25. 

WRITTEN  EXERCISE 

1.  Make  a  table  of  the  reciprocals  of  the  integers  from 
1  to  10,  using  the  inverted  slide. 

2.  Make  a  table  of  the  reciprocals  of  numbers  1.5,  1.6, 
1.7,  etc.,  to  2.5  inclusive. 

265.  Combined  multiplication  and  division. 

"24.5x46.5 

EXAMPLE  1.   Find  the  value  of ^-77— 

ou^U 

To  2.45  on  7)  set  3.62  on  C.  At  4.65  on  C  find  the  signifi- 
cant figures  of  the  result  315  on  D.  To  obtain  the  decimal 
point, 

24.5  x  46.5  =  2.45  x  10  X  4.65  x  10 
3620  3.62  x  1000 

2.45  x  4.65     _!_ 
3.62        X 10 
=  3.15  x  TV 
=  .315. 

EXAMPLE  2.   Find  the  value  of  -  — — - 

O.Zi 

To  1.46  on  D  set  5.2  on  C.  3.25  lies  beyond  the  extrem- 
ity of  D.  Bring  the  runner  to  the  index  10  on  C,  and  move 
the  slide  till  the  other  index  coincides  with  the  runner. 
This  operation  multiplies  the  result  by  10  and  does  not 


268  SHOP  PROBLEMS  IN  MATHEMATICS 

affect  the  significant  figures.  Under  3.25  on  C  read  the  re- 
sult 913  on  D.  The  position  of  the  decimal  point  may  be 
determined  by  the  method  of  Example  1  or  by  making  a 
rough  calculation  as  follows  : 

1.46  x  3.25  _  1.5  X  3 
5.2  5 

From  this  we  see  that  the  result  should  be  913. 
EXAMPLE  3.    Find  the  value  of 

24.4  x  580  x  36.8  x  9.65  x  1.4 
3.55  x  8450  x  5.15  x  1.24 

At  2.44  on  D  set  3.55  on  C;  runner  to  5.8  on  C  ;  8.45  on 
C  to  the  runner;  runner  to  3.68  on  C;  5.15  to  the  runner  ; 
runner  to  9.65  on  C  ;  1.24  on  C  to  the  runner.  At  1.4  on  C 
read  the  significant  figures  of  the  result  367. 

For  the  position  of  the  decimal  point,  making  a  rough 
calculation, 

8 
20  x          X       X  10  X 


x  l 


From  this  we  see  that  the  result  is  36.7. 

_,.  „  32.6  x  51.5  x  945 

EXAMPLE  4.   Find  the  value  of 


At  3.26  on  D  set  2.74  on  C  ;  runner  to  5.15  on  C  ;  4.8  on 
C  to  the  runner  ;  9.45  projects  beyond  T)  ;  runner  to  1  on 
C  ;  10  on  C  to  runner  ;  runner  to  9.45  on  C  ;  5.25  on  C  to 
the  runner. 

Under  the  index  10  of  C  read  the  significant  figures  of 
the  result  22.9. 

,  ,      30  x  50  x  900 
Calculating  roughly,  3  x  500  x  5Q  -  18. 

Hence  the  result  is  22.9. 


THE  SLIDE  RULE  269 

It  is  evident  that  any  series  of  multiplications  and  divi- 
sions may  be  reduced  to  the  form  of  Example  3  or  4  by  in- 
troducing the  factor  1  in  numerator  or  denominator.  Thus 

546  x  3.14  x  7.18  =  546  x  3.14  x  7.18 
9.12  9.12  x  1 

ORAL  EXERCISE 
Find  the  value  of : 

2.18  x  .0136  1 


5.44  '  3.66  x  .33  x  6.25 

7.85  x  1.25  2 

2.  ,-.,..  „  * . 


23.3  '  3.42  x  8.65 

574 


3.  28.6  x  5.8  x  7.14.        8. 


264  x  436 


2.34  x  4.82  3.14  x  5.6  x  .49 


9.38  x  1.36  '  25.3  x  .06  x  .685 

8.45  6.8  x  4.5  x  7.7  x  122 

5-  4.65  x  .0176 '  8.3  x  5.2  x  26  x  12 

266.  Proportion.    Problems  in  proportion  are  special  cases 
of  multiplication  and  division. 

Thus  a\b  =  c:x. 

x  =  --  (See  sect.  185.) 

Set  a  on  C  to  b  on  D,  and  opposite  c  on  C  find  x  on  D. 
Note  that  the  first  and  third  terms  of  the  proportion  are 
on  one  scale,  while  the  second  and  fourth  terms  are  on  the 
other  scale. 

EXAMPLE  1.   Solve  16  :  25  =  18.5 :  x. 

25  x  18.5 

T6~ 
Solve  by  the  method  of  section  265. 


270  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  2.    Solve  x  :  25  =  16.5  : 18. 

25  x  1.65 
x  = — 

Solve  as  before. 

ORAL  EXERCISE 
Find  x  in  the  following  proportions :    .  • 

1.  25  :  35  : :  18.5  :  x.  6.  67.5  :  59.4  : :  94.9  :  x, 

2.  1.5  :  2.6  :  :  13  :  x.  7.  81.7  :  x  : :  48.4  :  23. 

3.  3.74  :  2.3  : :  52.1  :  x.  8.  6.7  :  5.3  : :  x  :  4.2. 

4.  2.65 :  4.82  : :  7.58  :  x.  9.  3  :  17  : :  x  :  10. 

5.  x  :  6.42  :  :  11.2  :  14.4.  10.  5  :  19.  :  :  x  :  10. 

267.  Centimeters  into  inches.    On  the  slide  rule,  by  simple 
division,  centimeters  may  be  changed  into  inches. 

ORAL  EXERCISE 

Convert  into  inches : 

(1  in.  =  2.54  centimeters.) 
1.2.18cm.    2.5.25cm.    3.7.99cm.    4.8.78cm.    5.4.65cm. 

268.  Squares.    Over  any  number  on  D  is  found  its  square 
on  A.    Thus,  setting  the  runner  to  2  on  D,  4  will  be  indi- 
cated on  A.    Conversely,  2  on  D  is  the   square  root   of 
4  on  A. 

CASE  1.  When  the  number  lies  between  1  and  10,  the 
runner  set  to  the  number  on  D  will  indicate  the  square  on  A. 

EXAMPLE  1.    Square  1.35. 

The  runner  set  to  1.35  on  D  indicates  1.82  011  .1. 

CASE  2.  When  the  number  does  not  lie  between  1  and 
10 :  (1)  change  the  number  to  one  that  is  between  1  and  10 
by  moving  the  decimal  point ;  (2)  square  by  Case  1 ;  and 
(3)  move  the  decimal  point  twice  as  many  places  as  it  was 
moved  in  (1)  and  in  the  opposite  direction. 


THE  SLIDE  RULE  271 

EXAMPLE  2.    Square  75. 

752  =  (7.5)2  x  102  =  56.25  x  100  =  5625. 
EXAMPLE  3.    Square  .00355. 
.00355  =  (3.55)2  x  (10" 3)2  =  12.6  x  10-6  =  .0000126. 

ORAL   EXERCISE 

Find  the  squares  of : 

1.  4.75.       3.  9.65.        5.  24.2.       7.  .0635.         9.  .0346. 

2.  1.14.       4.  376.         6.  .56.         8.  .0087.       10.  2250. 

11.  Find  the  area  of  a  square  field  measuring  126  rd.  on 
a  side. 

12.  How  many  square  yards  are  there  in  a  circular  plot 
of  ground  measuring  12.5  yd.  in  diameter  ? 

The  area  of  a  circle  may  be  found  by  multiplying  TT  by  the  square 
of  the  radius,  or  by  multiplying  .7854,  the  area  of  a  circle  1  yd.  in 
diameter,  by  the  square  of  the  given  diameter. 

269.  Square  root.  Square  root  is  the  inverse  operation  of 
squaring  ;  hence  the  operations  with  the  slide  rule  must 
be  reversed. 

CASE  1.  When  the  number  lies  between  1  and  100,  the 
runner  set  to  the  number  on  A  will  indicate  the  square 
root  on  D. 


EXAMPLE  1.    Find  V38.5. 

The  runner  set  to  38.5  on  A  indicates  6.2,  the  square 
root,  on  D. 

CASE  2.  When  the  number  does  not  lie  between  1  and 
100  :  (1)  change  it  to  one  lying  between  1  and  100  by 
moving  the  decimal  point  an  even  number  of  places  ;  (2)  find 
the  square  root  by  Case  1 ;  and  (3)  move  the  decimal  point 
one  half  as  many  places  as  it  was  moved  in  (1)  and  in  the 
opposite  direction. 


272  SHOP  PROBLEMS  IN  MATHEMATICS 

EXAMPLE  2.    Find  Vl22. 

Vl22  =  VL22  x  VlOO  =  1.104  x  10  =  11.04 


EXAMPLE  3.    Find  Vl220. 

V1220  =  Vl^2  x  VlOO  =  3.49  x.10  =  34.9. 
EXAMPLE  4.    Find  V.00122. 

V.00122  =  VK2  x  VTF1  =  3.49  x  10- 2  =  .0349. 

ORAL  EXERCISE 

Find  the  square  roots  of : 

1.  48.5.       3.  485.         5.  2.58.       7.  .375.  9.  .0066. 

2.  4850.      4.  .0485.      6.  97.5.       8.  .31416.     10.  1540. 

11.  How  long  must  one  side  of  a  square  field  be  to  con- 
tain 5  A.  ? 

160  sq.  rd.  =  1  A. 

12.  Find    the    diameter    of   a    circular   cylinder   30"   in 
height  that  will  contain  5  gal. 

231  cu.  in.  =  1  gal. 

REVIEW   EXERCISE 
Find  the  value  of : 

5.35  x  425  x  (64.5)2 

.45  x  .072 
Use  scale  C  for  the  factor  that  is  to  be  squared. 

175  x  V31  x  VT8  x  250 
29  x  13  x  841 

Use  scale  B  for  the  square  roots. 

3.  14.6  xV2.  4.  8  V3. 


THE  SLIDE  RULE  273 

15.5  (.0247)2 

V3580'  "  (-0068)2' 

6      2'1  9.    Vl4  x  13  x  11  X  3. 

'  (8.4)* 

(18.4)2  x  56.2.  V640  x  1.54 

(3.7)2  124  x. 163   ' 


11.  The  area  of  a  triangle  is  V*  (s  —  a)  (s  —  b)  (s  —  c\ 
where  s  is  %  the  perimeter,  and  a,  b,  and  c  are  the  sides. 
If  a  =  4:7  rd.,  b  =  31  rd.,  and  c  =  22  rd.,  find  the  area  of 
the  field. 

12.  Using  the  slide  rule,  make  a  list  of  squares  of  inte- 
gral numbers  between  20  and  30. 

13.  Make  a  list  of  square  roots  of  integral  numbers  be- 
tween 70  and  80,  correct  to  one  place  of  decimals. 


TABLES 


TABLE   I 
TABLE  OF  NATURAL  SINES,  TANGENTS,  COSINES,  AND  COTANGENTS 


Degrees 

Sine 

Tangent 

Cotangent 

Cosine 

0 

0 

0 

8 

1 

90 

1 

2 
3 
4 

.0175 
.0349 
.0523 
.0698 

.0175 
.0349 
.0524 
.0699 

57.2900 
28.6363 
19.0811 
14.3007 

.  .  .9998 
.9994 
.9986 
.9976 

89 

88 
87 
86 

5 

.0872 

.0875 

11.4301 

.9962 

85 

6 

7 
8 
9 

.1045 
.1219 
.1392 
.1564 

.1051 
.1228 
.1405 
.1584 

9.5144 
8.1443 
7.1154 
6.3138 

.9945 
.9925 
.9903 

.9877 

84 
83 
82 
81 

10 

.1736 

.1763 

5.6713 

.9848 

80 

11 
12 

13 
14 

.1908 
.2079 
.2250 
.2419 

.1944 
.2126 
.2309 
.2493 

5.1446 
4.7046 
4.3315 
4.0108 

.9816 
.9781 
.9744 
.9703 

79 

78 
77 
76 

15 

.2588 

.2679 

3.7321 

.9659 

75 

16 
17 
18 
19 

.2756 
.2924 
.3090 
.3256 

.2867 
.3057 
.3249 
.3443 

3.4874 
3.2709 
3.0777 
2.9042 

.9613 
.9563 
.9511 
.9455 

74 
73 
72 
71 

20 

.3420 

.3640 

2.7475 

.9397 

70 

21 

22 
23 
24 

.3584 
.3746 
.3907 
.4067 

.3839 
.4040 
.4245 
.4452 

2.6051 
2.4751 
2.3559 
2.2460 

.9336 
.9272 
.9205 
.9135 

69 
68 
67 
66 

25 

.4226 

.4663 

2.1445 

.9063 

65 

26 
27 
28 
29 

.4384 
.4540 
.4695 
.4848 

.4877 
.5095 
.5317 
.5543 

2.0503 
1.9626 
1.8807 
1.8040 

.8988 
.8910 
.8829 
.8746 

64 

63 
62 
61 

30 

.5000 

.5774 

1.7321 

.8660 

60 

31 
32 
33 
34 

.5150 
.5299 
.5446 
.5592 

.6009 
.6249 
.6494 
.6745 

1.6643 
1.6003 
1.5399 
1.4826 

.8572 
.8480 
.8387 
.8290 

59 
58 
57 
56 

35 

.5736 

.7002 

1.4281 

.8192 

55 

36 
37 
38 
39 

.5878 
.6018 
.6157 
,6293 

.7265 
.7536 
.7813 
.8098 

1.3764 
1.3270 
1.2799 
1.2349 

.8090 
.7986 
.7880 
.7771 

54 
53 
52 
51 

40 

.6428 

.8391 

1.1918 

.7660 

50 

41 
42 
43 
44 

.6561 
.6691 
.6820 
.6947 

.8693 
.9004 
.9325 
.9657 

1.1504 
1.1106 
1.0724 
1.0355 

.7547 
.7431 
.7314 
.7193 

49 

48 
47 
46 

45 

.7071 

1.0000 

1.0000 

.7071 

45 

Cosine 

Cotangent 

Tangent 

Sine 

Degrees 

274 


TABLES 


275 


TABLE  II — WEIGHT  OF  CASTINGS  FROM  WEIGHT  OF  PATTERNS 

Factors  by  which  to  multiply  weight  of  pattern  to  get  weight  of  casting 


Material  of  pattern 

For  cast  iron 

For  brass 

For  copper 

For  aluminum 

White  pine     .    .    . 

16 
12 

l"l 

1? 

4| 
4* 

Mahogany  .... 
Brass 

U9* 

13* 
1 

U) 

Dry  sand    .... 
Aluminum      .     .    . 

? 

5 
3.4 

5 
3J 

U 

1 

TABLE  III  —  WEIGHTS  OF  MATERIALS 


Woods 

Lbs.  per  cu.  ft. 

Woods 

Lbs.  per  cu.  ft. 

Ash     
Birch      .     . 

52 
38 

Lignum  vitae      .     . 
Linden  

83.3 

377 

Cherry 

44 

Mahogany 

35  to  53 

Chestnut     

35 
28 

Maple    
Oak  (white) 

45 
50 

Elm    
Gum 

41 
40 

Oak  (red)  .... 
Pine  (white) 

52 
30 

Hemlock    .         ... 

24 

Pine  (yellow) 

43 

Hickory      

45  to  53 

Metals 

Metals 

A  1 

160 

Steel  .    ."    . 

499 

Babbit  Metal 

457 

Tin 

4555 

488 

Zinc  

437 

Copper   

545 

Other  material 

Cast  iron     
Wrought  iron     .     .     . 
Lead       

450 
486 
708.5 

Water   
Granite      .... 
Marble  
Brick  and  mortar  . 

62J 
169 
166 
120. 

TABLE  IV  — COMMON  NAILS 


Size 

Length 

Approximate 
No.  to  Ib. 

Size 

Length 

Approximate 
No.  to  Ib. 

2d 
3d 
4d 
5d 
6d 
7d 
8d 
9d 

1    inch 
1£  inch 
1£  inch 
If  inch 
2    inch 
2£  inch 
2£  inch 
2|  inch 

876 
568 
316 
271 
181 
161 
106 
96 

lOd 
12d 
16d 
20d 
30d 
40d 
50d 
60d 

3    inch 
3i  inch 
3|  inch 
4   inch 
4J  inch 
5    inch 
5£  inch 
6    inch 

69 
63 
49 
31 
24 
18 
14 
11 

TABLE  V  —  MISCELLANY 


A  gallon  contains 
A  bushel  contains 


231  cu.  in. 
2150  cu.  in. 


A  gallon  of  water  weighs 
A  barrel  contains  .     .     . 


8.35  Ib. 
31i  gal. 


276 


SHOP  PROBLEMS  IN  MATHEMATICS 


TABLE  VI 
FOR  FINDING  THE  UNKNOWN  PARTS  OF  GEARS  FROM  THE  KNOWN  l 


To  get 

Having 

Formula    , 

The  diametral  pitch 

The  circular  pitch  

p      3.1416 

The  diametral  pitch 
The  diametral  pitch 

The  pitch  diameter  and  the  number 
of  teeth    ». 

The  outside  diameter  and  the  number 

Po 

*-S 

p,  N+2 

Pitch  diameter    .     . 

of  teeth   

The  number  of  teeth  and  the  diame- 
tral pitch     

d~o« 

D=K 

Pitch  diameter    .    . 

The  number  of  teeth  and  the  outside 

ODxN 

Pitch  diameter    .    . 

Pitch  diameter    .     . 
Outside  diameter     . 

The  outside  diameter  and  the  diame- 
tral pitch     

Addendum  and  the  number  of  teeth  . 
The  number  of  teeth  and  the  diame- 

D=OI>- — 
Pd 

D=sN 

"N"  +  •> 
n~n     i>  T  <i 

Outside  diameter 

tral  pitch     

The  pitch  diameter  and  the  diametral 
pitch 

~    Pa 
OD-D  +  I- 

Outside  diameter 

The  pitch  diameter  and  the  number 
of  teeth 

OD-S^ 

Outside  diameter 
Number  of  teeth  .     . 

The  number  of  teeth  and  addendum  . 

The  pitch  diameter  and  the  diametral 
pitch                  .         .     .         .    . 

D 

OD=(N  +  2)  s 

N=DPa 

Number  of  teeth  .     . 

The  outside  diameter  and  the  diame- 
tral pitch     

N=OI)xPd-2 
1.5708 

The  diametral  pitch            .... 

Pd 
s-  1 

Root         

The  diametral  pitch  

Pd 
B  +  f-i^Z 

Working  depth    .    . 
Whole  depth    .     .     . 
Clearance    .... 
Clearance     .... 

The  diametral  pitch  
The  diametral  pitch  
The  diametral  pitch  
Thickness  of  tooth      

Pd 

*-i; 

w+i-8£ 

.157 

"ST 

f-  * 

"10 

l  Adapted  from  Nuttall. 


TABLES 


277 


TABLE  VII 


SIXES  OF  TAP  DRILLS  FOR  U.S. 
STANDARD  THREADS 

SIXES  OF  TAP  DRILLS  FOR  COMMON 
V-THREADS 

Outside 
diameter 
of  screw 
in  inches 

Threads 
per  inch 

Diameter 
of 
tap  drill 

Outside 
diameter 
of  screw 
in  inches 

Threads 
per  inch 

Diameter 
of 
tap  drill 

1 

1 

20 

11 

20 

18 

4 

4 

5 

5 

16 

18 

D 

16 

18 

2 

3 

19 

3 

8 

16 

64 

8 

16 

J 

7 

7 

16 

14 

S 

16 

14 

P 

1 

2 

13 

13 
32 

1 

2 

12 

U 

9 
16 

12 

15 
32 

9 
16 

12 

.428 

5 

8 

11 

33 
64 

5 
8 

11 

31 
64 

3 
4 

10 

5 

8 

11 
16 

11 

35 
64 

7 

3 

3 

19 

8 

9 

4 

4 

10 

3~2 

1 

8 

27 

13 

10 

21 

32 

16 

32 

H 

7 

61 
64 

7 
8 

9 

45 
64 

l| 

7 

1J 

15 

9 

49 

16 

64 

if 

6 

*H 

51 

i 

H 

6 

ifV 

j. 

64 

if 

6| 

ill 

if 

5 

11 

i| 

5 

lj 

2 

g 

ill 

Tap-drill  sizes  as  given  in  these  tables  are  from  five  to  ten  thousandths 
larger  than  the  actual  root  diameter  of  the  screw. 


278 


SHOP  PROBLEMS  IN  MATHEMATICS 


TABLE  VIII 
DECIMAL  EQUIVALENTS  OF  FRACTIONAL  SIZES 


1 

64 

.0156 

17 
64 

.2656 

33 
64 

.6156 

49 
64 

.7656 

1 
32 

.0312 

9 
32 

.2812 

17 
32 

.5312 

25 
32 

.7812 

3 
64 

.0468 

19 
64 

.2968 

35 
64 

.5468 

51 
64 

.7968 

1 
16 

.0625 

5 
16 

.3125 

9 
16 

.5625 

13 
16 

.8125 

5 
64 

.0781 

21 
64 

.3281 

37 
64 

.5781 

53 
64 

.8281 

3 
32 

.0937 

11 

32 

.3437 

19 
32 

.5937 

27 
32 

.8437 

7 

.1093 

23 

.3593 

39 

.6093 

55 

.8593 

64 

64 

64 

64 

1 

.125 

3 

.375 

5 

.625 

7 

.875 

8 

8 

8 

8 

9 

.1406 

25 

.3906 

41 

.6406 

57 

.8906 

64 

64 

64 

64 

5 
32 

.1562 

13 
32 

.4062 

21 
32 

.6562 

29 
32 

.9062 

11 
64 

.1718 

27 
64 

.4218 

43 
64 

.6718 

59 
64 

.9218 

3 
16 

.1875 

7 
16 

.4375 

11 
16 

.6875 

15 
16 

.9375 

13 
64 

.2031 

29 
64 

.4531 

45 
64 

.7031 

61 
64 

.9531 

7 
32 

.2187 

15 
32 

.4687 

23 

32 

.7187 

31 
32 

.9687 

15 

64 

.2343 

31 
64 

.4843 

47 
64 

.7343 

63 
64 

.9843 

1 

.250 

1 

.500 

3 

.750 

1 

1.000 

4 

2 

4 

INDEX 


Allowance,  in  cutting  logs,  3 

for  dressing,  2 

for   matching  and   waste   in 
flooring,  17 

for  shrinkage,  87 
Arc,  length  of,  221 
Averages,  195 
Back  gears,  111 
Barrels,  84 

Basket-ball  court,  202 
Belting,  coils  of,  72 
Belts,  lengths  of,  68-72 
Bicycles,  gear,  212 
Bins,  capacities  of,  41 
Blast  furnace,  98 
Brickwork,  36 
Bushel  measure,  74 
Casks,  84 
Castings,  87 
Chimneys,  37 
Chuck,  hollow,  90 
Circle,  220 

Circles  equivalent  to  two  circles,49 
Cisterns,  78 

Clearance  on  milling  cutters,  183 
Coal,  weight  in  cubic  feet,  42 

cubic  feet  in  ton,  43 
Coefficient,  238 
Cone,  231 

Cone  pulley,  91,  112 
Core  print,  88 
Cope,  92 

Cost  of  material,  98,  101 
Countershaft,  111 
Crank,  single-  and  double-throw, 

104 

Cross  slide,  142-144 
Cube,  225 

Cubical  contents,  approximate,  41 
Cupola,  94,  95,  97 
Cutting  speed,  119,  128,  129 
Cylinder,  228,  229 


Decimals,  190 

Distance  across  flats,  219 

across  river,  57 
Drag,  92 
Dressed,  1 

Drive,  direct  and  back-geared,  111 
Eccentric,  179 

Estimating  work,  98,  99,  123-128 
Exponent,  238 
Face,  18 
Face  plate,  90 
Feed,  121 
Floor  joists,  25,  26 
Flooring,  17 
Forgings,  102,  103,  106 
Formula,  definition,  13 
Formulas,  solution  of,  237 
Foundry,  87 
Frame  of  house,  26 
Framework,  24 
Gable  roof,  28 
Gambrel  roof,  30 
Gas  engine,  184 
Gearing,  compound,  158 

simple,  155 
Gear  teeth,  proportions  of,  166, 

167 
Gears,  114-118 

spur,  spiral,  and  bevel,  166 
Golden  section,  211 
Grade,  206,  207 
Heaps,  84 
Heights  of  cliffs,  58 
Hexagon,  distance  across  corners, 

176 

Hip  rafter,  30 
Hip  roof,  30 

Hogshead,  gallons  in,  85 
Hopper,  44,  75 
Index  head,  161 
Jack  rafter,  30,  31 
Jig  plates,  177-179 


279 


280 


SHOP  PROBLEMS  IN  MATHEMATICS 


Jointed,  definition,  3 
Lathe,  engine,  148 

wood-turning,  61 
Lathing,  38 
Lean-to  roof,  29 
Lumber,  3 
Machine  work  (cut  of  exercises), 

126,  127 

Metal,  weights,  87  (Table  III) 
Milling  machine  (cut  and  parts), 

174,  175 

Molds,  pressure  in,  92 
Mortar,  common  and  patent,  38 
Nosing,  21,  23 
Patterns,  87 
Perimeter,  215 
Pipe,  standard  sizes,  50 
Pitch,  diametral  and  circular,  166 

of  a  roof,  28-29 
Planer,  2,  130 
Plastering,  38 
Plate,  26,  28,  30,  31 
Polygon,  219 

Potatoes,  weights  per  bushel,  43 
Prism,  224 

Pulley,  driving  and  driven,  62 
Pulleys,  speed  of,  111-113 
Pyramid,  226 
Rafters,  length,  28 
Reciprocal,  189 
Rectangle,  213 
Return  bend,  88 
Rhombus,  228 
Ridge  of  roof,  30,  31 
Ring,  81,  82,  234 
Rise  of  rafter,  28,  30 

of  stairs,  21 
Riser,  21,  23 
Roof  work,  28 
Rough  stock,  1 
Rougher  lumber,  ordering,  24 
Run  of  rafter,  28,  30 

of  stairs,  21 

of  step,  21 
Saw,  band,  66 

circular,  67 

Screw  cutting,  gearing  for,  155-1 58 
Screws,  see  Threads 


Sector,  222 

Segment,  222 

Shaper,  cut  of,  120 

Sheet-metal  work,  74 

Shingles,  standard,  33 

Shingling,  33 

Shrinkage,  allowance,  87 

Sill,  25-26 

Slide  rests,  142,  143 

Slide  rule,  260 

Span  of  roof,  28 

Speed,  cutting  and  surface,  62 

Sphere,  233 

Spirals  on  drills  and  gears,  180, 

181,  182 
Sprue,  92 
Square  of  flooring,  19 

diagonal,  in  terms  of  side,  173 

equivalent  to  two  squares,  48 
Square  root,  197 
Stairs,  21 

Standard  lengths  of  lumber,  6 
Steel,  proportion  of  elements,  108 
Steps,  22 
Stonework,  41 
String,  23 
Studding,  27 
Surface    measure,    approximate 

method,  14 
Surface  plate,  89 
Surfaced,  3 
Tapers,  137-139 

Threads,  double  and  triple,  154, 
164 

metric,  159 

pitch,  lead,  depth,  147 

standard,  149-155 
Timber,  3,  4 
Trapezoid,  217 
Tread,  22,  23 
Triangle,  215,  216 

30°  and  45°,  223 
Valley  rafter,  31 
Valley  roof,  31 
Vernier  of  the  micrometer,  136, 

137 
Zone,  234 


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